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http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #PowerShell | PowerShell | $i = 0
foreach ($s in $args) {
Write-Host Argument (++$i) is $s
} |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Fortran | Fortran | C This would be some kind of comment
C Usually one would avoid columns 2-6 even in a comment. |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #FreeBASIC | FreeBASIC | ' FB 1.05.0 Win64
' This a single line comment
REM This is another way of writing a single line comment
/'
This is a
multi-line
comment
'/
/'
Multi-line comments
/'
can also be nested
'/
like this
'/ |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #Maxima | Maxima | life(A) := block(
[p, q, B: zerofor(A), s],
[p, q]: matrix_size(A),
for i thru p do (
for j thru q do (
s: 0,
if j > 1 then s: s + A[i, j - 1],
if j < q then s: s + A[i, j + 1],
if i > 1 then (
s: s + A[i - 1, j],
if j > 1 then s: s + A[i - 1, j - 1],
if j < q then s: s + A[i - 1, j + 1]
),
if i < p then (
s: s + A[i + 1, j],
if j > 1 then s: s + A[i + 1, j - 1],
if j < q then s: s + A[i + 1, j + 1]
),
B[i, j]: charfun(s = 3 or (s = 2 and A[i, j] = 1))
)
),
B
)$
/* a glider */
L: matrix([0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])$
gen(A, n) := block(thru n do A: life(A), A)$
gen(L, 4);
matrix([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]) |
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #F.23 | F# |
printfn "%s" (if 3<2 then "3 is less than 2" else "3 is not less than 2")
|
http://rosettacode.org/wiki/Compare_a_list_of_strings | Compare a list of strings | Task
Given a list of arbitrarily many strings, show how to:
test if they are all lexically equal
test if every string is lexically less than the one after it (i.e. whether the list is in strict ascending order)
Each of those two tests should result in a single true or false value, which could be used as the condition of an if statement or similar.
If the input list has less than two elements, the tests should always return true.
There is no need to provide a complete program and output.
Assume that the strings are already stored in an array/list/sequence/tuple variable (whatever is most idiomatic) with the name strings, and just show the expressions for performing those two tests on it (plus of course any includes and custom functions etc. that it needs), with as little distractions as possible.
Try to write your solution in a way that does not modify the original list, but if it does then please add a note to make that clear to readers.
If you need further guidance/clarification, see #Perl and #Python for solutions that use implicit short-circuiting loops, and #Raku for a solution that gets away with simply using a built-in language feature.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Vlang | Vlang | fn all_equal(strings []string) bool {
for s in strings {
if s != strings[0] {
return false
}
}
return true
}
fn all_less_than(strings []string) bool {
for i := 1; i < strings.len(); i++ {
if !(strings[i - 1] < s) {
return false
}
}
return true
} |
http://rosettacode.org/wiki/Compare_a_list_of_strings | Compare a list of strings | Task
Given a list of arbitrarily many strings, show how to:
test if they are all lexically equal
test if every string is lexically less than the one after it (i.e. whether the list is in strict ascending order)
Each of those two tests should result in a single true or false value, which could be used as the condition of an if statement or similar.
If the input list has less than two elements, the tests should always return true.
There is no need to provide a complete program and output.
Assume that the strings are already stored in an array/list/sequence/tuple variable (whatever is most idiomatic) with the name strings, and just show the expressions for performing those two tests on it (plus of course any includes and custom functions etc. that it needs), with as little distractions as possible.
Try to write your solution in a way that does not modify the original list, but if it does then please add a note to make that clear to readers.
If you need further guidance/clarification, see #Perl and #Python for solutions that use implicit short-circuiting loops, and #Raku for a solution that gets away with simply using a built-in language feature.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Wren | Wren | import "/sort" for Sort
var areEqual = Fn.new { |strings|
if (strings.count < 2) return true
return (1...strings.count).all { |i| strings[i] == strings[i-1] }
}
var areAscending = Fn.new { |strings| Sort.isSorted(strings) }
var a = ["a", "a", "a"]
var b = ["a", "b", "c"]
var c = ["a", "a", "b"]
var d = ["a", "d", "c"]
System.print("%(a) are all equal : %(areEqual.call(a))")
System.print("%(b) are ascending : %(areAscending.call(b))")
System.print("%(c) are all equal : %(areEqual.call(c))")
System.print("%(d) are ascending : %(areAscending.call(d))") |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #OCaml | OCaml | open Printf
let quibble list =
let rec aux = function
| a :: b :: c :: d :: rest -> a ^ ", " ^ aux (b :: c :: d :: rest)
| [a; b; c] -> sprintf "%s, %s and %s}" a b c
| [a; b] -> sprintf "%s and %s}" a b
| [a] -> sprintf "%s}" a
| [] -> "}" in
"{" ^ aux list
let test () =
[[];
["ABC"];
["ABC"; "DEF"];
["ABC"; "DEF"; "G"; "H"]]
|> List.iter (fun list -> print_endline (quibble list)) |
http://rosettacode.org/wiki/Combinations_with_repetitions | Combinations with repetitions | The set of combinations with repetitions is computed from a set,
S
{\displaystyle S}
(of cardinality
n
{\displaystyle n}
), and a size of resulting selection,
k
{\displaystyle k}
, by reporting the sets of cardinality
k
{\displaystyle k}
where each member of those sets is chosen from
S
{\displaystyle S}
.
In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter.
For example:
Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e.,
S
{\displaystyle S}
is
{
i
c
e
d
,
j
a
m
,
p
l
a
i
n
}
{\displaystyle \{\mathrm {iced} ,\mathrm {jam} ,\mathrm {plain} \}}
,
|
S
|
=
3
{\displaystyle |S|=3}
, and
k
=
2
{\displaystyle k=2}
.)
A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.
Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets.
Also note that doughnut can also be spelled donut.
Task
Write a function/program/routine/.. to generate all the combinations with repetitions of
n
{\displaystyle n}
types of things taken
k
{\displaystyle k}
at a time and use it to show an answer to the doughnut example above.
For extra credit, use the function to compute and show just the number of ways of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.
References
k-combination with repetitions
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #TXR | TXR | txr -p "(rcomb '(iced jam plain) 2)" |
http://rosettacode.org/wiki/Combinations_with_repetitions | Combinations with repetitions | The set of combinations with repetitions is computed from a set,
S
{\displaystyle S}
(of cardinality
n
{\displaystyle n}
), and a size of resulting selection,
k
{\displaystyle k}
, by reporting the sets of cardinality
k
{\displaystyle k}
where each member of those sets is chosen from
S
{\displaystyle S}
.
In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter.
For example:
Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e.,
S
{\displaystyle S}
is
{
i
c
e
d
,
j
a
m
,
p
l
a
i
n
}
{\displaystyle \{\mathrm {iced} ,\mathrm {jam} ,\mathrm {plain} \}}
,
|
S
|
=
3
{\displaystyle |S|=3}
, and
k
=
2
{\displaystyle k=2}
.)
A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.
Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets.
Also note that doughnut can also be spelled donut.
Task
Write a function/program/routine/.. to generate all the combinations with repetitions of
n
{\displaystyle n}
types of things taken
k
{\displaystyle k}
at a time and use it to show an answer to the doughnut example above.
For extra credit, use the function to compute and show just the number of ways of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.
References
k-combination with repetitions
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #Ursala | Ursala | #import std
#import nat
cwr = ~&s+ -<&*+ ~&K0=>&^|DlS/~& iota # takes a set and a selection size
#cast %gLSnX
main = ^|(~&,length) cwr~~/(<'iced','jam','plain'>,2) ('1234567890',3) |
http://rosettacode.org/wiki/Compiler/lexical_analyzer | Compiler/lexical analyzer | Definition from Wikipedia:
Lexical analysis is the process of converting a sequence of characters (such as in a computer program or web page) into a sequence of tokens (strings with an identified "meaning"). A program that performs lexical analysis may be called a lexer, tokenizer, or scanner (though "scanner" is also used to refer to the first stage of a lexer).
Task[edit]
Create a lexical analyzer for the simple programming language specified below. The
program should read input from a file and/or stdin, and write output to a file and/or
stdout. If the language being used has a lexer module/library/class, it would be great
if two versions of the solution are provided: One without the lexer module, and one with.
Input Specification
The simple programming language to be analyzed is more or less a subset of C. It supports the following tokens:
Operators
Name
Common name
Character sequence
Op_multiply
multiply
*
Op_divide
divide
/
Op_mod
mod
%
Op_add
plus
+
Op_subtract
minus
-
Op_negate
unary minus
-
Op_less
less than
<
Op_lessequal
less than or equal
<=
Op_greater
greater than
>
Op_greaterequal
greater than or equal
>=
Op_equal
equal
==
Op_notequal
not equal
!=
Op_not
unary not
!
Op_assign
assignment
=
Op_and
logical and
&&
Op_or
logical or
¦¦
The - token should always be interpreted as Op_subtract by the lexer. Turning some Op_subtract into Op_negate will be the job of the syntax analyzer, which is not part of this task.
Symbols
Name
Common name
Character
LeftParen
left parenthesis
(
RightParen
right parenthesis
)
LeftBrace
left brace
{
RightBrace
right brace
}
Semicolon
semi-colon
;
Comma
comma
,
Keywords
Name
Character sequence
Keyword_if
if
Keyword_else
else
Keyword_while
while
Keyword_print
print
Keyword_putc
putc
Identifiers and literals
These differ from the the previous tokens, in that each occurrence of them has a value associated with it.
Name
Common name
Format description
Format regex
Value
Identifier
identifier
one or more letter/number/underscore characters, but not starting with a number
[_a-zA-Z][_a-zA-Z0-9]*
as is
Integer
integer literal
one or more digits
[0-9]+
as is, interpreted as a number
Integer
char literal
exactly one character (anything except newline or single quote) or one of the allowed escape sequences, enclosed by single quotes
'([^'\n]|\\n|\\\\)'
the ASCII code point number of the character, e.g. 65 for 'A' and 10 for '\n'
String
string literal
zero or more characters (anything except newline or double quote), enclosed by double quotes
"[^"\n]*"
the characters without the double quotes and with escape sequences converted
For char and string literals, the \n escape sequence is supported to represent a new-line character.
For char and string literals, to represent a backslash, use \\.
No other special sequences are supported. This means that:
Char literals cannot represent a single quote character (value 39).
String literals cannot represent strings containing double quote characters.
Zero-width tokens
Name
Location
End_of_input
when the end of the input stream is reached
White space
Zero or more whitespace characters, or comments enclosed in /* ... */, are allowed between any two tokens, with the exceptions noted below.
"Longest token matching" is used to resolve conflicts (e.g., in order to match <= as a single token rather than the two tokens < and =).
Whitespace is required between two tokens that have an alphanumeric character or underscore at the edge.
This means: keywords, identifiers, and integer literals.
e.g. ifprint is recognized as an identifier, instead of the keywords if and print.
e.g. 42fred is invalid, and neither recognized as a number nor an identifier.
Whitespace is not allowed inside of tokens (except for chars and strings where they are part of the value).
e.g. & & is invalid, and not interpreted as the && operator.
For example, the following two program fragments are equivalent, and should produce the same token stream except for the line and column positions:
if ( p /* meaning n is prime */ ) {
print ( n , " " ) ;
count = count + 1 ; /* number of primes found so far */
}
if(p){print(n," ");count=count+1;}
Complete list of token names
End_of_input Op_multiply Op_divide Op_mod Op_add Op_subtract
Op_negate Op_not Op_less Op_lessequal Op_greater Op_greaterequal
Op_equal Op_notequal Op_assign Op_and Op_or Keyword_if
Keyword_else Keyword_while Keyword_print Keyword_putc LeftParen RightParen
LeftBrace RightBrace Semicolon Comma Identifier Integer
String
Output Format
The program output should be a sequence of lines, each consisting of the following whitespace-separated fields:
the line number where the token starts
the column number where the token starts
the token name
the token value (only for Identifier, Integer, and String tokens)
the number of spaces between fields is up to you. Neatly aligned is nice, but not a requirement.
This task is intended to be used as part of a pipeline, with the other compiler tasks - for example:
lex < hello.t | parse | gen | vm
Or possibly:
lex hello.t lex.out
parse lex.out parse.out
gen parse.out gen.out
vm gen.out
This implies that the output of this task (the lexical analyzer) should be suitable as input to any of the Syntax Analyzer task programs.
Diagnostics
The following error conditions should be caught:
Error
Example
Empty character constant
''
Unknown escape sequence.
\r
Multi-character constant.
'xx'
End-of-file in comment. Closing comment characters not found.
End-of-file while scanning string literal. Closing string character not found.
End-of-line while scanning string literal. Closing string character not found before end-of-line.
Unrecognized character.
|
Invalid number. Starts like a number, but ends in non-numeric characters.
123abc
Test Cases
Input
Output
Test Case 1:
/*
Hello world
*/
print("Hello, World!\n");
4 1 Keyword_print
4 6 LeftParen
4 7 String "Hello, World!\n"
4 24 RightParen
4 25 Semicolon
5 1 End_of_input
Test Case 2:
/*
Show Ident and Integers
*/
phoenix_number = 142857;
print(phoenix_number, "\n");
4 1 Identifier phoenix_number
4 16 Op_assign
4 18 Integer 142857
4 24 Semicolon
5 1 Keyword_print
5 6 LeftParen
5 7 Identifier phoenix_number
5 21 Comma
5 23 String "\n"
5 27 RightParen
5 28 Semicolon
6 1 End_of_input
Test Case 3:
/*
All lexical tokens - not syntactically correct, but that will
have to wait until syntax analysis
*/
/* Print */ print /* Sub */ -
/* Putc */ putc /* Lss */ <
/* If */ if /* Gtr */ >
/* Else */ else /* Leq */ <=
/* While */ while /* Geq */ >=
/* Lbrace */ { /* Eq */ ==
/* Rbrace */ } /* Neq */ !=
/* Lparen */ ( /* And */ &&
/* Rparen */ ) /* Or */ ||
/* Uminus */ - /* Semi */ ;
/* Not */ ! /* Comma */ ,
/* Mul */ * /* Assign */ =
/* Div */ / /* Integer */ 42
/* Mod */ % /* String */ "String literal"
/* Add */ + /* Ident */ variable_name
/* character literal */ '\n'
/* character literal */ '\\'
/* character literal */ ' '
5 16 Keyword_print
5 40 Op_subtract
6 16 Keyword_putc
6 40 Op_less
7 16 Keyword_if
7 40 Op_greater
8 16 Keyword_else
8 40 Op_lessequal
9 16 Keyword_while
9 40 Op_greaterequal
10 16 LeftBrace
10 40 Op_equal
11 16 RightBrace
11 40 Op_notequal
12 16 LeftParen
12 40 Op_and
13 16 RightParen
13 40 Op_or
14 16 Op_subtract
14 40 Semicolon
15 16 Op_not
15 40 Comma
16 16 Op_multiply
16 40 Op_assign
17 16 Op_divide
17 40 Integer 42
18 16 Op_mod
18 40 String "String literal"
19 16 Op_add
19 40 Identifier variable_name
20 26 Integer 10
21 26 Integer 92
22 26 Integer 32
23 1 End_of_input
Test Case 4:
/*** test printing, embedded \n and comments with lots of '*' ***/
print(42);
print("\nHello World\nGood Bye\nok\n");
print("Print a slash n - \\n.\n");
2 1 Keyword_print
2 6 LeftParen
2 7 Integer 42
2 9 RightParen
2 10 Semicolon
3 1 Keyword_print
3 6 LeftParen
3 7 String "\nHello World\nGood Bye\nok\n"
3 38 RightParen
3 39 Semicolon
4 1 Keyword_print
4 6 LeftParen
4 7 String "Print a slash n - \\n.\n"
4 33 RightParen
4 34 Semicolon
5 1 End_of_input
Additional examples
Your solution should pass all the test cases above and the additional tests found Here.
Reference
The C and Python versions can be considered reference implementations.
Related Tasks
Syntax Analyzer task
Code Generator task
Virtual Machine Interpreter task
AST Interpreter task
| #Zig | Zig |
const std = @import("std");
pub const TokenType = enum {
unknown,
multiply,
divide,
mod,
add,
subtract,
negate,
less,
less_equal,
greater,
greater_equal,
equal,
not_equal,
not,
assign,
bool_and,
bool_or,
left_paren,
right_paren,
left_brace,
right_brace,
semicolon,
comma,
kw_if,
kw_else,
kw_while,
kw_print,
kw_putc,
identifier,
integer,
string,
eof,
// More efficient implementation can be done with `std.enums.directEnumArray`.
pub fn toString(self: @This()) []const u8 {
return switch (self) {
.unknown => "UNKNOWN",
.multiply => "Op_multiply",
.divide => "Op_divide",
.mod => "Op_mod",
.add => "Op_add",
.subtract => "Op_subtract",
.negate => "Op_negate",
.less => "Op_less",
.less_equal => "Op_lessequal",
.greater => "Op_greater",
.greater_equal => "Op_greaterequal",
.equal => "Op_equal",
.not_equal => "Op_notequal",
.not => "Op_not",
.assign => "Op_assign",
.bool_and => "Op_and",
.bool_or => "Op_or",
.left_paren => "LeftParen",
.right_paren => "RightParen",
.left_brace => "LeftBrace",
.right_brace => "RightBrace",
.semicolon => "Semicolon",
.comma => "Comma",
.kw_if => "Keyword_if",
.kw_else => "Keyword_else",
.kw_while => "Keyword_while",
.kw_print => "Keyword_print",
.kw_putc => "Keyword_putc",
.identifier => "Identifier",
.integer => "Integer",
.string => "String",
.eof => "End_of_input",
};
}
};
pub const TokenValue = union(enum) {
intlit: i32,
string: []const u8,
};
pub const Token = struct {
line: usize,
col: usize,
typ: TokenType = .unknown,
value: ?TokenValue = null,
};
// Error conditions described in the task.
pub const LexerError = error{
EmptyCharacterConstant,
UnknownEscapeSequence,
MulticharacterConstant,
EndOfFileInComment,
EndOfFileInString,
EndOfLineInString,
UnrecognizedCharacter,
InvalidNumber,
};
pub const Lexer = struct {
content: []const u8,
line: usize,
col: usize,
offset: usize,
start: bool,
const Self = @This();
pub fn init(content: []const u8) Lexer {
return Lexer{
.content = content,
.line = 1,
.col = 1,
.offset = 0,
.start = true,
};
}
pub fn buildToken(self: Self) Token {
return Token{ .line = self.line, .col = self.col };
}
pub fn buildTokenT(self: Self, typ: TokenType) Token {
return Token{ .line = self.line, .col = self.col, .typ = typ };
}
pub fn curr(self: Self) u8 {
return self.content[self.offset];
}
// Alternative implementation is to return `Token` value from `next()` which is
// arguably more idiomatic version.
pub fn next(self: *Self) ?u8 {
// We use `start` in order to make the very first invocation of `next()` to return
// the very first character. It should be possible to avoid this variable.
if (self.start) {
self.start = false;
} else {
const newline = self.curr() == '\n';
self.offset += 1;
if (newline) {
self.col = 1;
self.line += 1;
} else {
self.col += 1;
}
}
if (self.offset >= self.content.len) {
return null;
} else {
return self.curr();
}
}
pub fn peek(self: Self) ?u8 {
if (self.offset + 1 >= self.content.len) {
return null;
} else {
return self.content[self.offset + 1];
}
}
fn divOrComment(self: *Self) LexerError!?Token {
var result = self.buildToken();
if (self.peek()) |peek_ch| {
if (peek_ch == '*') {
_ = self.next(); // peeked character
while (self.next()) |ch| {
if (ch == '*') {
if (self.peek()) |next_ch| {
if (next_ch == '/') {
_ = self.next(); // peeked character
return null;
}
}
}
}
return LexerError.EndOfFileInComment;
}
}
result.typ = .divide;
return result;
}
fn identifierOrKeyword(self: *Self) !Token {
var result = self.buildToken();
const init_offset = self.offset;
while (self.peek()) |ch| : (_ = self.next()) {
switch (ch) {
'_', 'a'...'z', 'A'...'Z', '0'...'9' => {},
else => break,
}
}
const final_offset = self.offset + 1;
if (std.mem.eql(u8, self.content[init_offset..final_offset], "if")) {
result.typ = .kw_if;
} else if (std.mem.eql(u8, self.content[init_offset..final_offset], "else")) {
result.typ = .kw_else;
} else if (std.mem.eql(u8, self.content[init_offset..final_offset], "while")) {
result.typ = .kw_while;
} else if (std.mem.eql(u8, self.content[init_offset..final_offset], "print")) {
result.typ = .kw_print;
} else if (std.mem.eql(u8, self.content[init_offset..final_offset], "putc")) {
result.typ = .kw_putc;
} else {
result.typ = .identifier;
result.value = TokenValue{ .string = self.content[init_offset..final_offset] };
}
return result;
}
fn string(self: *Self) !Token {
var result = self.buildToken();
result.typ = .string;
const init_offset = self.offset;
while (self.next()) |ch| {
switch (ch) {
'"' => break,
'\n' => return LexerError.EndOfLineInString,
'\\' => {
switch (self.peek() orelse return LexerError.EndOfFileInString) {
'n', '\\' => _ = self.next(), // peeked character
else => return LexerError.UnknownEscapeSequence,
}
},
else => {},
}
} else {
return LexerError.EndOfFileInString;
}
const final_offset = self.offset + 1;
result.value = TokenValue{ .string = self.content[init_offset..final_offset] };
return result;
}
/// Choose either `ifyes` or `ifno` token type depending on whether the peeked
/// character is `by`.
fn followed(self: *Self, by: u8, ifyes: TokenType, ifno: TokenType) Token {
var result = self.buildToken();
if (self.peek()) |ch| {
if (ch == by) {
_ = self.next(); // peeked character
result.typ = ifyes;
} else {
result.typ = ifno;
}
} else {
result.typ = ifno;
}
return result;
}
/// Raise an error if there's no next `by` character but return token with `typ` otherwise.
fn consecutive(self: *Self, by: u8, typ: TokenType) LexerError!Token {
const result = self.buildTokenT(typ);
if (self.peek()) |ch| {
if (ch == by) {
_ = self.next(); // peeked character
return result;
} else {
return LexerError.UnrecognizedCharacter;
}
} else {
return LexerError.UnrecognizedCharacter;
}
}
fn integerLiteral(self: *Self) LexerError!Token {
var result = self.buildTokenT(.integer);
const init_offset = self.offset;
while (self.peek()) |ch| {
switch (ch) {
'0'...'9' => _ = self.next(), // peeked character
'_', 'a'...'z', 'A'...'Z' => return LexerError.InvalidNumber,
else => break,
}
}
const final_offset = self.offset + 1;
result.value = TokenValue{
.intlit = std.fmt.parseInt(i32, self.content[init_offset..final_offset], 10) catch {
return LexerError.InvalidNumber;
},
};
return result;
}
// This is a beautiful way of how Zig allows to remove bilerplate and at the same time
// to not lose any error completeness guarantees.
fn nextOrEmpty(self: *Self) LexerError!u8 {
return self.next() orelse LexerError.EmptyCharacterConstant;
}
fn integerChar(self: *Self) LexerError!Token {
var result = self.buildTokenT(.integer);
switch (try self.nextOrEmpty()) {
'\'', '\n' => return LexerError.EmptyCharacterConstant,
'\\' => {
switch (try self.nextOrEmpty()) {
'n' => result.value = TokenValue{ .intlit = '\n' },
'\\' => result.value = TokenValue{ .intlit = '\\' },
else => return LexerError.EmptyCharacterConstant,
}
switch (try self.nextOrEmpty()) {
'\'' => {},
else => return LexerError.MulticharacterConstant,
}
},
else => {
result.value = TokenValue{ .intlit = self.curr() };
switch (try self.nextOrEmpty()) {
'\'' => {},
else => return LexerError.MulticharacterConstant,
}
},
}
return result;
}
};
pub fn lex(allocator: std.mem.Allocator, content: []u8) !std.ArrayList(Token) {
var tokens = std.ArrayList(Token).init(allocator);
var lexer = Lexer.init(content);
while (lexer.next()) |ch| {
switch (ch) {
' ' => {},
'*' => try tokens.append(lexer.buildTokenT(.multiply)),
'%' => try tokens.append(lexer.buildTokenT(.mod)),
'+' => try tokens.append(lexer.buildTokenT(.add)),
'-' => try tokens.append(lexer.buildTokenT(.subtract)),
'<' => try tokens.append(lexer.followed('=', .less_equal, .less)),
'>' => try tokens.append(lexer.followed('=', .greater_equal, .greater)),
'=' => try tokens.append(lexer.followed('=', .equal, .assign)),
'!' => try tokens.append(lexer.followed('=', .not_equal, .not)),
'(' => try tokens.append(lexer.buildTokenT(.left_paren)),
')' => try tokens.append(lexer.buildTokenT(.right_paren)),
'{' => try tokens.append(lexer.buildTokenT(.left_brace)),
'}' => try tokens.append(lexer.buildTokenT(.right_brace)),
';' => try tokens.append(lexer.buildTokenT(.semicolon)),
',' => try tokens.append(lexer.buildTokenT(.comma)),
'&' => try tokens.append(try lexer.consecutive('&', .bool_and)),
'|' => try tokens.append(try lexer.consecutive('|', .bool_or)),
'/' => {
if (try lexer.divOrComment()) |token| try tokens.append(token);
},
'_', 'a'...'z', 'A'...'Z' => try tokens.append(try lexer.identifierOrKeyword()),
'"' => try tokens.append(try lexer.string()),
'0'...'9' => try tokens.append(try lexer.integerLiteral()),
'\'' => try tokens.append(try lexer.integerChar()),
else => {},
}
}
try tokens.append(lexer.buildTokenT(.eof));
return tokens;
}
pub fn main() !void {
var arena = std.heap.ArenaAllocator.init(std.heap.page_allocator);
defer arena.deinit();
const allocator = arena.allocator();
var arg_it = std.process.args();
_ = try arg_it.next(allocator) orelse unreachable; // program name
const file_name = arg_it.next(allocator);
// We accept both files and standard input.
var file_handle = blk: {
if (file_name) |file_name_delimited| {
const fname: []const u8 = try file_name_delimited;
break :blk try std.fs.cwd().openFile(fname, .{});
} else {
break :blk std.io.getStdIn();
}
};
defer file_handle.close();
const input_content = try file_handle.readToEndAlloc(allocator, std.math.maxInt(usize));
const tokens = try lex(allocator, input_content);
const pretty_output = try tokenListToString(allocator, tokens);
_ = try std.io.getStdOut().write(pretty_output);
}
fn tokenListToString(allocator: std.mem.Allocator, token_list: std.ArrayList(Token)) ![]u8 {
var result = std.ArrayList(u8).init(allocator);
var w = result.writer();
for (token_list.items) |token| {
const common_args = .{ token.line, token.col, token.typ.toString() };
if (token.value) |value| {
const init_fmt = "{d:>5}{d:>7} {s:<15}";
switch (value) {
.string => |str| _ = try w.write(try std.fmt.allocPrint(
allocator,
init_fmt ++ "{s}\n",
common_args ++ .{str},
)),
.intlit => |i| _ = try w.write(try std.fmt.allocPrint(
allocator,
init_fmt ++ "{d}\n",
common_args ++ .{i},
)),
}
} else {
_ = try w.write(try std.fmt.allocPrint(allocator, "{d:>5}{d:>7} {s}\n", common_args));
}
}
return result.items;
}
|
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Pure | Pure |
using system;
printf "There are %d command line argumants\n" argc;
puts "They are " $$ map (puts) argv;
|
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #PureBasic | PureBasic | If OpenConsole()
Define n=CountProgramParameters()
PrintN("Reading all parameters")
While n
PrintN(ProgramParameter())
n-1
Wend
Print(#CRLF$+"Press Enter")
Input()
CloseConsole()
EndIf |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Frink | Frink |
// This is a single-line comment
/* This is a comment
that spans multiple lines
and so on.
*/
|
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Futhark | Futhark |
-- Single-line comment
-- Multi-line
-- comment (yes, just several single-line comments).
|
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #MiniScript | MiniScript | // Conway's Game of Life
clear
rows = 64; rowRange = range(0, rows-1)
cols = 96; colRange = range(0, cols-1)
// prepare two tile displays, in display layers 4 and 5
img = Image.create(2, 1); img.setPixel 1, 0, color.white
grids = []
for dispIdx in [4,5]
display(dispIdx).mode = displayMode.tile
td = display(dispIdx)
td.cellSize = 10 // size of cells on screen
td.extent = [cols, rows]
td.overlap = -1 // adds a small gap between cells
td.tileSet = img; td.tileSetTileSize = 1
td.clear 0
grids.push td
end for
// initialize to a random state
for x in colRange
for y in rowRange
td.setCell x, y, rnd > 0.5
end for
end for
curIdx = 5
// main loop
while true
yield
td = grids[curIdx-4]
newIdx = 4 + (curIdx == 4)
newTd = grids[newIdx-4]
for x in colRange
for y in rowRange
// get sum of neighboring states
sum = 0
for i in [x-1, x, x+1]
if i < 0 or i >= cols then continue
for j in [y-1, y, y+1]
if j < 0 or j >= rows then continue
if i==x and j==y then continue
sum = sum + td.cell(i,j)
end for
end for
// Now, update the cell based on current state
// and neighboring sum.
if td.cell(x,y) then
newTd.setCell x, y, (sum == 2 or sum == 3)
else
newTd.setCell x, y, sum == 3
end if
end for
end for
display(newIdx).mode = displayMode.tile
display(curIdx).mode = displayMode.off
curIdx = newIdx
end while |
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #Factor | Factor |
t 1 2 ? ! returns 1
|
http://rosettacode.org/wiki/Compare_a_list_of_strings | Compare a list of strings | Task
Given a list of arbitrarily many strings, show how to:
test if they are all lexically equal
test if every string is lexically less than the one after it (i.e. whether the list is in strict ascending order)
Each of those two tests should result in a single true or false value, which could be used as the condition of an if statement or similar.
If the input list has less than two elements, the tests should always return true.
There is no need to provide a complete program and output.
Assume that the strings are already stored in an array/list/sequence/tuple variable (whatever is most idiomatic) with the name strings, and just show the expressions for performing those two tests on it (plus of course any includes and custom functions etc. that it needs), with as little distractions as possible.
Try to write your solution in a way that does not modify the original list, but if it does then please add a note to make that clear to readers.
If you need further guidance/clarification, see #Perl and #Python for solutions that use implicit short-circuiting loops, and #Raku for a solution that gets away with simply using a built-in language feature.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #XProfan | XProfan | Proc allsame
Parameters long liste
var int result = 1
var int cnt = GetCount(liste)
Case cnt == 0 : Return 0
Case cnt == 1 : Return 1
WhileLoop 1, cnt-1
If GetString$(liste,&loop - 1) <> GetString$(liste,&loop)
result = 0
BREAK
EndIf
EndWhile
Return result
EndProc
Proc strict_order
Parameters long liste
var int result = 1
var int cnt = GetCount(liste)
Case cnt == 0 : Return 0
Case cnt == 1 : Return 1
WhileLoop 1, cnt-1
If GetString$(liste,&loop) <= GetString$(liste,&loop - 1)
result = 0
BREAK
EndIf
EndWhile
Return result
EndProc
cls
declare string s[4]
s[0] = "AA,BB,CC"
s[1] = "AA,AA,AA"
s[2] = "AA,CC,BB"
s[3] = "AA,ACB,BB,CC"
s[4] = "single_element"
WhileLoop 0,4
ClearList 0
Move("StrToList",s[&loop],",")
Print "list:",s[&loop]
Print "...is " + if(allsame(0), "", "not ") + "lexically equal"
Print "...is " + if(strict_order(0), "", "not ") + "in strict ascending order"
EndWhile
ClearList 0
WaitKey
end |
http://rosettacode.org/wiki/Compare_a_list_of_strings | Compare a list of strings | Task
Given a list of arbitrarily many strings, show how to:
test if they are all lexically equal
test if every string is lexically less than the one after it (i.e. whether the list is in strict ascending order)
Each of those two tests should result in a single true or false value, which could be used as the condition of an if statement or similar.
If the input list has less than two elements, the tests should always return true.
There is no need to provide a complete program and output.
Assume that the strings are already stored in an array/list/sequence/tuple variable (whatever is most idiomatic) with the name strings, and just show the expressions for performing those two tests on it (plus of course any includes and custom functions etc. that it needs), with as little distractions as possible.
Try to write your solution in a way that does not modify the original list, but if it does then please add a note to make that clear to readers.
If you need further guidance/clarification, see #Perl and #Python for solutions that use implicit short-circuiting loops, and #Raku for a solution that gets away with simply using a built-in language feature.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #zkl | zkl | fcn allEQ(strings){ (not strings.filter1('!=(strings[0]))) }
fcn monoUp(strings){
strings.len()<2 or
strings.reduce(fcn(a,b){ if(a>=b) return(Void.Stop,False); b }).toBool()
} |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #Oforth | Oforth | : quibbing(l) -- string
| i s |
StringBuffer new "{" <<
l size dup 1- ->s loop: i [
l at(i) <<
i s < ifTrue: [ ", " << continue ]
i s == ifTrue: [ " and " << ]
]
"}" << dup freeze ; |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #Ol | Ol |
(define (quibble . args)
(display "{")
(let loop ((args args))
(unless (null? args) (begin
(display (car args))
(cond
((= 1 (length args)) #t)
((= 2 (length args))
(display " and "))
(else
(display ", ")))
(loop (cdr args)))))
(print "}"))
; testing =>
(quibble)
(quibble "ABC")
(quibble "ABC" "DEF")
(quibble "ABC" "DEF" "G" "H")
|
http://rosettacode.org/wiki/Combinations_with_repetitions | Combinations with repetitions | The set of combinations with repetitions is computed from a set,
S
{\displaystyle S}
(of cardinality
n
{\displaystyle n}
), and a size of resulting selection,
k
{\displaystyle k}
, by reporting the sets of cardinality
k
{\displaystyle k}
where each member of those sets is chosen from
S
{\displaystyle S}
.
In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter.
For example:
Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e.,
S
{\displaystyle S}
is
{
i
c
e
d
,
j
a
m
,
p
l
a
i
n
}
{\displaystyle \{\mathrm {iced} ,\mathrm {jam} ,\mathrm {plain} \}}
,
|
S
|
=
3
{\displaystyle |S|=3}
, and
k
=
2
{\displaystyle k=2}
.)
A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.
Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets.
Also note that doughnut can also be spelled donut.
Task
Write a function/program/routine/.. to generate all the combinations with repetitions of
n
{\displaystyle n}
types of things taken
k
{\displaystyle k}
at a time and use it to show an answer to the doughnut example above.
For extra credit, use the function to compute and show just the number of ways of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.
References
k-combination with repetitions
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #VBScript | VBScript | ' Combinations with repetitions - iterative - VBScript
Sub printc(vi,n,vs)
Dim i, w
For i=0 To n-1
w=w &" "& vs(vi(i))
Next 'i
Wscript.Echo w
End Sub
Sub combine(flavors, draws, xitem, tell)
Dim n, i, j
ReDim v(draws)
If tell Then Wscript.Echo "list of cwr("& flavors &","& draws &") :"
Do While True
For i=0 To draws-1
If v(i)>flavors-1 Then
v(i+1)=v(i+1)+1
For j=i To 0 Step -1
v(j)=v(i+1)
Next 'j
End If
Next 'i
If v(draws)>0 Then Exit Do
n=n+1
If tell Then Call printc(v, draws, xitem)
v(0)=v(0)+1
Loop
Wscript.Echo "cwr("& flavors &","& draws &")="&n
End Sub
items=Array( "iced", "jam", "plain" )
combine 3, 2, items, True
combine 10, 3, , False
combine 10, 7, , False
combine 10, 9, , False |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Python | Python | import sys
program_name = sys.argv[0]
arguments = sys.argv[1:]
count = len(arguments) |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #R | R | R CMD BATCH --vanilla --slave '--args a=1 b=c(2,5,6)' test.r test.out |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #FutureBasic | FutureBasic |
// Single line comment
' Single line comment
rem Single line comment
/* Single line comment */
/*
Multiline
comment
*/
|
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #FUZE_BASIC | FUZE BASIC | //Comment (No space required)
# Comment (Space required)
REM Comment (Space require)
PRINT "This is an inline comment."//Comment (No space required)
END |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #Nim | Nim | import os, strutils, random
randomize()
var w, h: int
if paramCount() >= 2:
w = parseInt(paramStr(1))
h = parseInt(paramStr(2))
if w <= 0: w = 30
if h <= 0: h = 30
# Initialize
var univ, utmp = newSeq[seq[bool]](h)
for y in 0..<h:
univ[y].newSeq w
utmp[y].newSeq w
for x in 0 ..< w:
if rand(9) < 1:
univ[y][x] = true
while true:
# Show
stdout.write "\e[H"
for y in 0..<h:
for x in 0..<w:
stdout.write if univ[y][x]: "\e[07m \e[m" else: " "
stdout.write "\e[E"
stdout.flushFile
# Evolve
for y in 0..<h:
for x in 0..<w:
var n = 0
for y1 in y-1..y+1:
for x1 in x-1..x+1:
if univ[(y1+h) mod h][(x1 + w) mod w]:
inc n
if univ[y][x]: dec n
utmp[y][x] = n == 3 or (n == 2 and univ[y][x])
swap(univ,utmp)
sleep 200 |
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #FALSE | FALSE | condition[body]? |
http://rosettacode.org/wiki/Compare_a_list_of_strings | Compare a list of strings | Task
Given a list of arbitrarily many strings, show how to:
test if they are all lexically equal
test if every string is lexically less than the one after it (i.e. whether the list is in strict ascending order)
Each of those two tests should result in a single true or false value, which could be used as the condition of an if statement or similar.
If the input list has less than two elements, the tests should always return true.
There is no need to provide a complete program and output.
Assume that the strings are already stored in an array/list/sequence/tuple variable (whatever is most idiomatic) with the name strings, and just show the expressions for performing those two tests on it (plus of course any includes and custom functions etc. that it needs), with as little distractions as possible.
Try to write your solution in a way that does not modify the original list, but if it does then please add a note to make that clear to readers.
If you need further guidance/clarification, see #Perl and #Python for solutions that use implicit short-circuiting loops, and #Raku for a solution that gets away with simply using a built-in language feature.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #zonnon | zonnon |
module CompareStrings;
type
Vector = array * of string;
var
v,w: Vector;
i: integer;
all,ascending: boolean;
begin
v := new Vector(3);
v[0] := "uno";
v[1] := "uno";
v[2] := "uno";
all := true;
for i := 1 to len(v) - 1 do
all := all & (v[i - 1] = v[i]);
end;
w := new Vector(3);
w[0] := "abc";
w[1] := "bcd";
w[2] := "cde";
v := w;
ascending := true;
for i := 1 to len(v) - 1 do
ascending := ascending & (v[i - 1] <= v[i])
end;
write("all equals?: ");writeln(all);
write("ascending?: ");writeln(ascending)
end CompareStrings.
|
http://rosettacode.org/wiki/Compare_a_list_of_strings | Compare a list of strings | Task
Given a list of arbitrarily many strings, show how to:
test if they are all lexically equal
test if every string is lexically less than the one after it (i.e. whether the list is in strict ascending order)
Each of those two tests should result in a single true or false value, which could be used as the condition of an if statement or similar.
If the input list has less than two elements, the tests should always return true.
There is no need to provide a complete program and output.
Assume that the strings are already stored in an array/list/sequence/tuple variable (whatever is most idiomatic) with the name strings, and just show the expressions for performing those two tests on it (plus of course any includes and custom functions etc. that it needs), with as little distractions as possible.
Try to write your solution in a way that does not modify the original list, but if it does then please add a note to make that clear to readers.
If you need further guidance/clarification, see #Perl and #Python for solutions that use implicit short-circuiting loops, and #Raku for a solution that gets away with simply using a built-in language feature.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #ZX_Spectrum_Basic | ZX Spectrum Basic | 10 FOR j=160 TO 200 STEP 10
20 RESTORE j
30 READ n
40 LET test1=1: LET test2=1
50 FOR i=1 TO n
60 READ a$
70 PRINT a$;" ";
80 IF i=1 THEN GO TO 110
90 IF p$<>a$ THEN LET test1=0
100 IF p$>=a$ THEN LET test2=0
110 LET p$=a$
120 NEXT i
130 PRINT 'test1'test2
140 NEXT j
150 STOP
160 DATA 3,"AA","BB","CC"
170 DATA 3,"AA","AA","AA"
180 DATA 3,"AA","CC","BB"
190 DATA 4,"AA","ACB","BB","CC"
200 DATA 1,"single_element" |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #PARI.2FGP | PARI/GP | comma(v)={
if(#v==0, return("{}"));
if(#v==1, return(Str("{"v[1]"}")));
my(s=Str("{",v[1]));
for(i=2,#v-1,s=Str(s,", ",v[i]));
Str(s," and ",v[#v],"}")
};
comma([])
comma(["ABC"])
comma(["ABC", "DEF"])
comma(["ABC", "DEF", "G", "H"]) |
http://rosettacode.org/wiki/Combinations_with_repetitions | Combinations with repetitions | The set of combinations with repetitions is computed from a set,
S
{\displaystyle S}
(of cardinality
n
{\displaystyle n}
), and a size of resulting selection,
k
{\displaystyle k}
, by reporting the sets of cardinality
k
{\displaystyle k}
where each member of those sets is chosen from
S
{\displaystyle S}
.
In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter.
For example:
Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e.,
S
{\displaystyle S}
is
{
i
c
e
d
,
j
a
m
,
p
l
a
i
n
}
{\displaystyle \{\mathrm {iced} ,\mathrm {jam} ,\mathrm {plain} \}}
,
|
S
|
=
3
{\displaystyle |S|=3}
, and
k
=
2
{\displaystyle k=2}
.)
A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.
Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets.
Also note that doughnut can also be spelled donut.
Task
Write a function/program/routine/.. to generate all the combinations with repetitions of
n
{\displaystyle n}
types of things taken
k
{\displaystyle k}
at a time and use it to show an answer to the doughnut example above.
For extra credit, use the function to compute and show just the number of ways of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.
References
k-combination with repetitions
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #Wren | Wren | var combrep // recursive
combrep = Fn.new { |n, lst|
if (n == 0 ) return [[]]
if (lst.count == 0) return []
var r = combrep.call(n, lst[1..-1])
for (x in combrep.call(n-1, lst)) {
var y = x.toList
y.add(lst[0])
r.add(y)
}
return r
}
System.print(combrep.call(2, ["iced", "jam", "plain"]))
System.print(combrep.call(3, (1..10).toList).count) |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Racket | Racket | #lang racket
(current-command-line-arguments) |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Raku | Raku | # with arguments supplied
$ raku -e 'sub MAIN($x, $y) { say $x + $y }' 3 5
8
# missing argument:
$ raku -e 'sub MAIN($x, $y) { say $x + $y }' 3
Usage:
-e '...' x y |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #RapidQ | RapidQ | PRINT "This program is named "; Command$(0)
FOR i=1 TO CommandCount
PRINT "The argument "; i; " is "; Command$(i)
NEXT i |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Gambas | Gambas |
' This whole line is a comment and is ignored by the gambas interpreter
print "Hello" ' Comments after an apostrophe are ignored
'' A bold-style comment
' TODO: To Do comment will appear in Task Bar
' FIXME: Fix Me comment will appear in Task Bar
' NOTE: Note commnet will appear in Task Bar
|
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #GAP | GAP | # Comment (till end of line) |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #OCaml | OCaml | let get g x y =
try g.(x).(y)
with _ -> 0
let neighbourhood g x y =
(get g (x-1) (y-1)) +
(get g (x-1) (y )) +
(get g (x-1) (y+1)) +
(get g (x ) (y-1)) +
(get g (x ) (y+1)) +
(get g (x+1) (y-1)) +
(get g (x+1) (y )) +
(get g (x+1) (y+1))
let next_cell g x y =
let n = neighbourhood g x y in
match g.(x).(y), n with
| 1, 0 | 1, 1 -> 0 (* lonely *)
| 1, 4 | 1, 5 | 1, 6 | 1, 7 | 1, 8 -> 0 (* overcrowded *)
| 1, 2 | 1, 3 -> 1 (* lives *)
| 0, 3 -> 1 (* get birth *)
| _ (* 0, (0|1|2|4|5|6|7|8) *) -> 0 (* barren *)
let copy g = Array.map Array.copy g
let next g =
let width = Array.length g
and height = Array.length g.(0)
and new_g = copy g in
for x = 0 to pred width do
for y = 0 to pred height do
new_g.(x).(y) <- (next_cell g x y)
done
done;
(new_g)
let print g =
let width = Array.length g
and height = Array.length g.(0) in
for x = 0 to pred width do
for y = 0 to pred height do
if g.(x).(y) = 0
then print_char '.'
else print_char 'o'
done;
print_newline()
done |
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #Fancy | Fancy | if: (x < y) then: {
"x < y!" println # will only execute this block if x < y
}
|
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #Pascal | Pascal |
program CommaQuibbling;
uses
SysUtils,
Classes,
StrUtils;
const
OuterBracket =['[', ']'];
type
{$IFNDEF FPC}
SizeInt = LongInt;
{$ENDIF}
{ TCommaQuibble }
TCommaQuibble = class(TStringList)
private
function GetCommaquibble: string;
procedure SetCommaQuibble(AValue: string);
public
property CommaQuibble: string read GetCommaquibble write SetCommaQuibble;
end;
{$IFNDEF FPC} // Delphi support
function WordPosition(const N: Integer; const S: string; const WordDelims:
TSysCharSet): SizeInt;
var
PS, P, PE: PChar;
Count: Integer;
begin
Result := 0;
Count := 0;
PS := PChar(pointer(S));
PE := PS + Length(S);
P := PS;
while (P < PE) and (Count <> N) do
begin
while (P < PE) and (P^ in WordDelims) do
inc(P);
if (P < PE) then
inc(Count);
if (Count <> N) then
while (P < PE) and not (P^ in WordDelims) do
inc(P)
else
Result := (P - PS) + 1;
end;
end;
function ExtractWordPos(N: Integer; const S: string; const WordDelims:
TSysCharSet; out Pos: Integer): string;
var
i, j, l: SizeInt;
begin
j := 0;
i := WordPosition(N, S, WordDelims);
if (i > High(Integer)) then
begin
Result := '';
Pos := -1;
Exit;
end;
Pos := i;
if (i <> 0) then
begin
j := i;
l := Length(S);
while (j <= l) and not (S[j] in WordDelims) do
inc(j);
end;
SetLength(Result, j - i);
if ((j - i) > 0) then
Result := copy(S, i, j - i);
end;
function ExtractWord(N: Integer; const S: string; const WordDelims: TSysCharSet):
string; inline;
var
i: SizeInt;
begin
Result := ExtractWordPos(N, S, WordDelims, i);
end;
{$ENDIF}
{ TCommaQuibble }
procedure TCommaQuibble.SetCommaQuibble(AValue: string);
begin
AValue := ExtractWord(1, AValue, OuterBracket);
commatext := AValue;
end;
function TCommaQuibble.GetCommaquibble: string;
var
x: Integer;
Del: string;
begin
result := '';
Del := ', ';
for x := 0 to Count - 1 do
begin
result := result + Strings[x];
if x = Count - 2 then
Del := ' and '
else if x = Count - 1 then
Del := '';
result := result + Del;
end;
result := '{' + result + '}';
end;
const
TestData: array[0..7] of string = ('[]', '["ABC"]', '["ABC", "DEF"]',
'["ABC", "DEF", "G", "H"]', '', '"ABC"', '"ABC", "DEF"', '"ABC", "DEF", "G", "H"');
var
Quibble: TCommaQuibble;
TestString: string;
begin
Quibble := TCommaQuibble.Create;
for TestString in TestData do
begin
Quibble.CommaQuibble := TestString;
writeln(Quibble.CommaQuibble);
end;
end. |
http://rosettacode.org/wiki/Combinations_with_repetitions | Combinations with repetitions | The set of combinations with repetitions is computed from a set,
S
{\displaystyle S}
(of cardinality
n
{\displaystyle n}
), and a size of resulting selection,
k
{\displaystyle k}
, by reporting the sets of cardinality
k
{\displaystyle k}
where each member of those sets is chosen from
S
{\displaystyle S}
.
In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter.
For example:
Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e.,
S
{\displaystyle S}
is
{
i
c
e
d
,
j
a
m
,
p
l
a
i
n
}
{\displaystyle \{\mathrm {iced} ,\mathrm {jam} ,\mathrm {plain} \}}
,
|
S
|
=
3
{\displaystyle |S|=3}
, and
k
=
2
{\displaystyle k=2}
.)
A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.
Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets.
Also note that doughnut can also be spelled donut.
Task
Write a function/program/routine/.. to generate all the combinations with repetitions of
n
{\displaystyle n}
types of things taken
k
{\displaystyle k}
at a time and use it to show an answer to the doughnut example above.
For extra credit, use the function to compute and show just the number of ways of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.
References
k-combination with repetitions
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #XPL0 | XPL0 | code ChOut=8, CrLf=9, IntOut=11, Text=12;
int Count, Array(10);
proc Combos(D, S, K, N, Names); \Generate all size K combinations of N objects
int D, S, K, N, Names; \depth of recursion, starting value of N, etc.
int I;
[if D<K then \depth < size
[for I:= S to N-1 do
[Array(D):= I;
Combos(D+1, I, K, N, Names);
];
]
else [Count:= Count+1;
if Names(0) then
[for I:= 0 to K-1 do
[Text(0, Names(Array(I))); ChOut(0, ^ )];
CrLf(0);
];
];
];
[Count:= 0;
Combos(0, 0, 2, 3, ["iced", "jam", "plain"]);
Text(0, "Combos = "); IntOut(0, Count); CrLf(0);
Count:= 0;
Combos(0, 0, 3, 10, [0]);
Text(0, "Combos = "); IntOut(0, Count); CrLf(0);
] |
http://rosettacode.org/wiki/Combinations_with_repetitions | Combinations with repetitions | The set of combinations with repetitions is computed from a set,
S
{\displaystyle S}
(of cardinality
n
{\displaystyle n}
), and a size of resulting selection,
k
{\displaystyle k}
, by reporting the sets of cardinality
k
{\displaystyle k}
where each member of those sets is chosen from
S
{\displaystyle S}
.
In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter.
For example:
Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e.,
S
{\displaystyle S}
is
{
i
c
e
d
,
j
a
m
,
p
l
a
i
n
}
{\displaystyle \{\mathrm {iced} ,\mathrm {jam} ,\mathrm {plain} \}}
,
|
S
|
=
3
{\displaystyle |S|=3}
, and
k
=
2
{\displaystyle k=2}
.)
A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.
Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets.
Also note that doughnut can also be spelled donut.
Task
Write a function/program/routine/.. to generate all the combinations with repetitions of
n
{\displaystyle n}
types of things taken
k
{\displaystyle k}
at a time and use it to show an answer to the doughnut example above.
For extra credit, use the function to compute and show just the number of ways of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.
References
k-combination with repetitions
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #zkl | zkl | fcn combosK(k,seq){ // repeats, seq is finite
if (k==1) return(seq);
if (not seq) return(T);
self.fcn(k-1,seq).apply(T.extend.fp(seq[0])).extend(self.fcn(k,seq[1,*]));
} |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Raven | Raven | ARGS print
stack (6 items)
0 => "raven"
1 => "myprogram"
2 => "-c"
3 => "alpha beta"
4 => "-h"
5 => "gamma" |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #REALbasic | REALbasic | Function Run(args() as String) As Integer
For each arg As String In args
Stdout.WriteLine(arg)
Next
End Function |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #REXX | REXX | say 'command arguments:'
say arg(1) |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #gecho | gecho | ( this is a test comment... o.O ) 1 2 + . |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Gema | Gema | ! comment starts with "!" and continues to end of line |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #OCTAVE | OCTAVE |
clear all
x=55; % Size of the Lattice (same as LAWE)
z(1:1:x^2)=0; % Initialise the binary lattice
z_prime=z; % prepare the z prime
idx=7*x+2; % Origin
z(idx+4)=1; % Populate the binary lattice with the Gosper Glider
z(idx+x+4)=1;
z(idx+1+4)=1;
z(idx+x+1+4)=1;
z(idx+14)=1;
z(idx+14+x)=1;
z(idx+14+x+x)=1;
z(idx+15+x+x+x)=1;
z(idx+15-x)=1;
z(idx+16-x-x)=1;
z(idx+16+x+x+x+x)=1;
z(idx+17-x-x)=1;
z(idx+17+x+x+x+x)=1;
z(idx+18+x)=1;
z(idx+19-x)=1;
z(idx+19+x+x+x)=1;
z(idx+20)=1;
z(idx+20+x)=1;
z(idx+20+x+x)=1;
z(idx+21+x)=1;
z(idx+24)=1;
z(idx+25)=1;
z(idx+24-x)=1;
z(idx+25-x)=1;
z(idx+24-x-x)=1;
z(idx+25-x-x)=1;
z(idx+26-x-x-x)=1;
z(idx+26+x)=1;
z(idx+28-x-x-x)=1;
z(idx+28+x)=1;
z(idx+28-x-x-x-x)=1;
z(idx+28+x+x)=1;
z(idx+38)=1;
z(idx+38-x)=1;
z(idx+39)=1;
z(idx-x+39)=1;
for k=1:1:80; % Number of frames
for n=x+2:1:x^2-x-1; % one time pass
theta=0; % Sum the surrounding area (top equation)
for c=n-1:1:n+1;
for m=-1:1:1;
theta=theta+z(m*x+c);
endfor
endfor
z_prime(n)= round(1/5*(z(n)*acos(sin(pi/4*(theta-z(n)-9/2))) + (1-z(n)) * acos(sin(pi/4*(theta-z(n) + 3))))); % bottom equation
endfor
figure(2)
title('GOL','FontWeight','bold');
xlabel("X");
ylabel("Y");
imagesc(reshape(z,x,x)')
z=z_prime;
pause % each press advances the algorithm one step
endfor
|
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #Forth | Forth | ( condition ) IF ( true statements ) THEN
( condition ) IF ( true statements ) ELSE ( false statements ) THEN |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #Perl | Perl | sub comma_quibbling(@) {
return "{$_}" for
@_ < 2 ? "@_" :
join(', ', @_[0..@_-2]) . ' and ' . $_[-1];
}
print comma_quibbling(@$_), "\n" for
[], [qw(ABC)], [qw(ABC DEF)], [qw(ABC DEF G H)]; |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #Phix | Phix | function quibble(sequence words)
if length(words)>=2 then
words[-2..-1] = {words[-2]&" and "&words[-1]}
end if
return "{"&join(words,", ")&"}"
end function
constant tests = {{},
{"ABC"},
{"ABC","DEF"},
{"ABC","DEF","G","H"}}
for i=1 to length(tests) do
?quibble(tests[i])
end for
|
http://rosettacode.org/wiki/Combinations_with_repetitions | Combinations with repetitions | The set of combinations with repetitions is computed from a set,
S
{\displaystyle S}
(of cardinality
n
{\displaystyle n}
), and a size of resulting selection,
k
{\displaystyle k}
, by reporting the sets of cardinality
k
{\displaystyle k}
where each member of those sets is chosen from
S
{\displaystyle S}
.
In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter.
For example:
Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e.,
S
{\displaystyle S}
is
{
i
c
e
d
,
j
a
m
,
p
l
a
i
n
}
{\displaystyle \{\mathrm {iced} ,\mathrm {jam} ,\mathrm {plain} \}}
,
|
S
|
=
3
{\displaystyle |S|=3}
, and
k
=
2
{\displaystyle k=2}
.)
A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.
Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets.
Also note that doughnut can also be spelled donut.
Task
Write a function/program/routine/.. to generate all the combinations with repetitions of
n
{\displaystyle n}
types of things taken
k
{\displaystyle k}
at a time and use it to show an answer to the doughnut example above.
For extra credit, use the function to compute and show just the number of ways of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.
References
k-combination with repetitions
See also
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #ZX_Spectrum_Basic | ZX Spectrum Basic | 10 READ n
20 DIM d$(n,5)
30 FOR i=1 TO n
40 READ d$(i)
50 NEXT i
60 DATA 3,"iced","jam","plain"
70 FOR i=1 TO n
80 FOR j=i TO n
90 PRINT d$(i);" ";d$(j)
100 NEXT j
110 NEXT i |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Ring | Ring |
see copy("=",30) + nl
see "Command Line Parameters" + nl
see "Size : " + len(sysargv) + nl
see sysargv
see copy("=",30) + nl
for x = 1 to len(sysargv)
see x + nl
next
|
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Ruby | Ruby | #! /usr/bin/env ruby
p ARGV |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Rust | Rust | use std::env;
fn main(){
let args: Vec<_> = env::args().collect();
println!("{:?}", args);
} |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Genie | Genie | // Comment continues until end of line
/* Comment lasts between delimiters */ |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #GML | GML | // comment starts with "//" and continues to the end of the line |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #Ol | Ol |
#!/usr/bin/ol
(import (otus random!))
(define MAX 65536) ; should be power of two
; size of game board (should be less than MAX)
(define WIDTH 170)
(define HEIGHT 96)
; helper function
(define (hash x y)
(let ((x (mod (+ x WIDTH) WIDTH))
(y (mod (+ y HEIGHT) HEIGHT)))
(+ (* y MAX) x)))
; helper function
(define neighbors '(
(-1 . -1) ( 0 . -1) ( 1 . -1)
(-1 . 0) ( 1 . 0)
(-1 . 1) ( 0 . 1) ( 1 . 1)
))
; dead-or-alive cell test
(define (alive gen x y)
(case (fold (lambda (f xy)
(+ f (get gen (hash (+ x (car xy)) (+ y (cdr xy))) 0)))
0 neighbors)
(2
(get gen (hash x y) #false))
(3
#true)))
; ---------------
(import (lib gl2))
(gl:set-window-title "Convey's The game of Life")
(glShadeModel GL_SMOOTH)
(glClearColor 0.11 0.11 0.11 1)
(glOrtho 0 WIDTH 0 HEIGHT 0 1)
(glPointSize (/ 854 WIDTH))
; generate random field
(gl:set-userdata
(list->ff (map (lambda (i) (let ((x (rand! WIDTH)) (y (rand! HEIGHT)))
(cons (hash x y) 1))) (iota 2000))))
; main game loop
(gl:set-renderer (lambda (mouse)
(let ((generation (gl:get-userdata)))
(glClear GL_COLOR_BUFFER_BIT)
; draw the cells
(glColor3f 0.2 0.5 0.2)
(glBegin GL_POINTS)
(ff-fold (lambda (st key value)
(glVertex2f (mod key MAX)
(div key MAX))
) #f generation)
(glEnd)
(gl:set-userdata
; next cells generation
(ff-fold (lambda (st key value)
(let ((x (mod key MAX))
(y (div key MAX)))
(fold (lambda (st key)
(let ((x (+ x (car key)))
(y (+ y (cdr key))))
(if (alive generation x y) (put st (hash x y) 1) st)))
(if (alive generation x y) (put st (hash x y) 1) st) ; the cell
neighbors)))
#empty generation)))))
|
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #Fortran | Fortran | if ( a .gt. 20.0 ) then
q = q + a**2
else if ( a .ge. 0.0 ) then
q = q + 2*a**3
else
q = q - a
end if |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #PHP | PHP | <?php
function quibble($arr){
$words = count($arr);
if($words == 0){
return '{}';
}elseif($words == 1){
return '{'.$arr[0].'}';
}elseif($words == 2){
return '{'.$arr[0].' and '.$arr[1].'}';
}else{
return '{'.implode(', ', array_splice($arr, 0, -1) ). ' and '.$arr[0].'}';
}
}
$tests = [
[],
["ABC"],
["ABC", "DEF"],
["ABC", "DEF", "G", "H"]
];
foreach ($tests as $test) {
echo quibble($test) . PHP_EOL;
} |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #S-lang | S-lang | variable a;
foreach a (__argv)
print(a);
|
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Sather | Sather | class MAIN is
main(args:ARRAY{STR}) is
loop
#OUT + args.elt! + "\n";
end;
end;
end; |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Scala | Scala | object CommandLineArguments extends App {
println(s"Received the following arguments: + ${args.mkString("", ", ", ".")}")
} |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #gnuplot | gnuplot | # this is a comment
# backslash continues \
a comment to the next \
line or lines |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Go | Go | // this is a single line comment
/* this is
a multi-line
block comment.
/* It does not nest */ |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #ooRexx | ooRexx | /* REXX ---------------------------------------------------------------
* 07.08.2014 Walter Pachl Conway's Game of life graphical
* Input is a file containing the initial pattern.
* The compute area is extended when needed
* (i.e., when cells are born outside the current compute area)
* When computing the pattern sequence is complete, the graphical
* output starts and continues until Cancel is pressed.
* 10.08.2014 WP fixed the output of what.txt
*--------------------------------------------------------------------*/
Parse Arg what speed
If what='?' Then Do
Say 'Create a file containing the pattern to be processed'
Say 'named somename.in (octagon.in such as this for the octagon):'
Say ' ** '
Say ' * * '
Say ' * * '
Say '* *'
Say '* *'
Say ' * * '
Say ' * * '
Say ' ** '
Say 'Run the program by entering "rexx conlife somename [pause]"',
'on the command line.'
Say '(pause is the amount of milliseconds between 2 pictures.',
'default is 1000)'
Say 'A file somename.lst will be created.'
Say 'Hereafter you will see the pattern''s development',
'in a new window.'
Say 'Press the Cancel button to end the presentation.'
Exit
End
Parse Version interpreter '_' level '('
If interpreter<>'REXX-ooRexx' Then Do
Say interpreter level
Say 'This program must be run with object Rexx.'
Exit
End
If what='' Then what='octagon'
If right(what,3)='.in' then
what=left(what,length(what)-3)
infile=what'.in'
If lines(infile)=0 Then Do
Say 'Input file' infile 'not found.'
Say 'Enter conlife ? for help.'
Exit
End
If speed='' Then
speed=1000
.local~myspeed=speed
Call tl what
--'type' what'.lst'
.local~title=what
array=.local~myarrayData
d = .drawDlg~new
if d~initCode <> 0 then do
say 'The Draw dialog was not created correctly. Aborting.'
return d~initCode
end
d~execute("SHOWTOP")
return 0
::requires "ooDialog.cls"
::class 'drawDlg' subclass UserDialog
::attribute interrupted unguarded
::method init
expose walterFont
forward class (super) continue
-- colornames:
-- 1 dark red 7 light grey 13 red
-- 2 dark green 8 pale green 14 light green
-- 3 dark yellow 9 light blue 15 yellow
-- 4 dark blue 10 white 16 blue
-- 5 purple 11 grey 17 pink
-- 6 blue grey 12 dark grey 18 turquoise
self~interrupted = .true
-- Create a font to write the nice big letters and digits
opts = .directory~new
opts~weight = 700
walterFont = self~createFontEx("Arial",14,opts)
walterFont = self~createFontEx("Courier",18,opts)
if \self~createcenter(200, 235,"Walter's Clock", , ,"System",14) then
self~initCode = 1
::method defineDialog
self~createPushButton(/*IDC_PB_DRAW*/100, 0, 0,240,200,"DISABLED NOTAB") -- The drawing surface.
self~createPushButton(IDCANCEL,160,220, 35, 12,,"&Cancel")
::method initDialog unguarded
expose x y dc myPen change al. fid nn what array
change = 0
x = self~factorx
y = self~factory
dc = self~getButtonDC(100)
myPen = self~createPen(1,'solid',0)
t = .TimeSpan~fromMicroSeconds(500000) -- .5 seconds
msg = .Message~new(self,'life')
alrm = .Alarm~new(t, msg)
array=.local~myArrayData
Do s=1 to array~items
al.s=array[s]
Parse Var al.s ' == ' al.s
End
nn=s-2
--say 'nn'nn
Call lineout fid
::method interrupt unguarded
self~interrupted = .true
::method cancel unguarded -- Stop the drawing program and quit.
expose x y
self~hide
self~interrupted = .true
return self~cancel:super
::method leaving unguarded -- Best place to clean up resources
expose dc myPen walterFont
self~deleteObject(myPen)
self~freeButtonDC(/*IDC_PB_DRAW*/100,dc)
self~deleteFont(walterFont)
::method life unguarded /* draw individual pixels */
expose x y dc myPen change walterFont al. nn what
mx = trunc(20*x); my = trunc(20*y); size = 400
curPen = self~objectToDC(dc, myPen)
-- Select the nice big letters and digits into the device context to use to
-- to write with:
curFont = self~fontToDC(dc, walterFont)
-- Create a white brush and select it into the device to paint with.
whiteBrush = self~createBrush(10)
curBrush = self~objectToDC(dc, whiteBrush)
-- Paint the drawing area surface with the white brush
self~rectangle(dc, 1, 1, 500, 600, 'FILL')
self~writeDirect(dc, 10, 20,'Conway''s Game of Life')
self~writeDirect(dc, 10, 40,.local~title)
self~writeDirect(dc, 10,460,'Walter Pachl, 8 Aug 2014')
dx=.local~dxval
dy=.local~dyval
do s=1 By 1 until self~interrupted
self~transparentText(dc)
self~interrupted = .false
sm=s//nn+1
If s>1 Then Do
ali=al.sb
Do While ali<>''
Parse Var ali x ',' y ali
zxa=(x+dx)*10
zya=(y+dy)*10
self~draw_square(dc,zxa,zya,3,10)
End
End
self~draw_square(dc, 380, 10,100,10)
self~writeDirect(dc, 340, 20,time())
self~writeDirect(dc, 340, 40,right(sm,2) 'of' right(nn,2))
ali=al.sm
Do While ali<>''
Parse Var ali x ',' y ali
zxa=(x+dx)*10
zya=(y+dy)*10
self~draw_square(dc,zxa,zya,3,5)
End
-- self~interrupted = .true
sb=sm
self~objectToDC(dc, curPen)
self~objectToDC(dc, curBrush)
call msSleep .local~myspeed
--self~pause
End
::method pause
j = msSleep(10)
::method draw_square
Use Arg dc, x, y, d, c
Do zx=x-d to x+d
Do zy=y-d to y+d
self~drawPixel(dc, zx, zy, c)
End
End
::method quot
Parse Arg x,y
If y=0 Then Return '???'
Else Return x/y
::routine tl
/* REXX ---------------------------------------------------------------
* 02.08.2014 Walter Pachl
* Input is a file containing the initial pattern
* The compute area is extended when needed
* (cells are born outside the current compute area)
* The program stops when the picture shown is the same as the first
* or equal to the previous one
*--------------------------------------------------------------------*/
Parse Arg f
If f='' Then f='bipole'
fid=f'.in'
oid=f'.txt'; 'erase' oid
oil=f'.lst'; 'erase' oil
debug=0
If debug Then Do
dbg=f'.xxx'; 'erase' dbg
End
ml=0
l.=''
ol.=''
Parse Value '10 10' With xb yb
xc=copies(' ',xb)
Do ri=yb+1 By 1 While lines(fid)>0
l.ri=xc||linein(fid)
ml=max(ml,length(strip(l.ri,'T')))
End
ri=ri-1
ml=ml+xb
ri=ri+yb
yy=ri
a.=' '
b.=' '
m.=''
x.=''
list.=''
Parse Value 1 ml 1 yy With xmi xma ymi yma
Parse Value '-10 30 -10 30' With xmi xma ymi yma
Parse Value '999 -999 999 -999 999 -999 999 -999',
With xmin xmax ymin ymax xlo xhi ylo yhi
Do y=1 To yy
z=yy-y+1
l=l.z
Do x=1 By 1 While l<>''
Parse Var l c +1 l
If c='*' Then
a.x.z='*'
End
End
Call show
Do step=1 To 60
Call store
If step>1 & is_equal(step,1) Then Leave
If step>1 & is_equal(step,step-1) Then Leave
Call show_neighbors
Do y=yma To ymi By -1
ol=format(x,3)' '
Do x=xmi To xma
neighbors=neighbors(x,y)
If a.x.y=' ' Then Do /* dead cell */
If neighbors=3 Then Do
b.x.y='*' /* gets life */
mmo=xmi xma ymi yma
xmi=min(xmi,x-1)
xma=max(xma,x+1)
ymi=min(ymi,y-1)
yma=max(yma,y+1)
mm=xmi xma ymi yma
If mm<>mmo Then
Call debug mmo '1->' mm
End
Else /* life cell */
b.x.y=' ' /* remains dead */
End
Else Do /* life cell */
If neighbors=2 |,
neighbors=3 Then Do
b.x.y='*' /* remains life */
mmo=xmi xma ymi yma
xmi=min(xmi,x-1)
xma=max(xma,x+1)
ymi=min(ymi,y-1)
yma=max(yma,y+1)
mm=xmi xma ymi yma
If mm<>mmo Then
Call debug mmo '2->' mm
End
Else
b.x.y=' ' /* dies */
End
End
End
/* b. is the new state and is now copied to a. */
Do y=yma To ymi By -1
Do x=xmi To xma
a.x.y=b.x.y
End
End
End
/* Output name and all states */
Call lineout oid,' 'f
st=' +' /* top and bottom border */
sb=' +' /* top and bottom border */
Do s=1 To step
st=st||'-'right(s,2,'-')||copies('-',xmax-xmin-2)'+'
sb=sb||copies('-',xmax-xmin+1)'+'
End
array=.array~new
Do y=ymin To ymax
Do s=1 To step
Do x=xmin To xmax
If substr(m.s.y,x,1)='*' Then Do
xlo=min(xlo,x)
xhi=max(xhi,x)
ylo=min(ylo,y)
yhi=max(yhi,y)
End
End
End
End
Do y=ymin To ymax
ol=''
Do s=1 To step
Do x=xmin To xmax
If substr(m.s.y,x,1)='*' Then Do
list.s=list.s (x-xlo+1)','||(y-ylo+1)
End
End
array[s]=s '-' words(list.s) '==' list.s
End
--Call lineout oid,ol '|'
.local~myArrayData=array
End
height=yhi-ylo+1
width=xhi-xlo+1
.local~dxval=(48-width)%2
.local~dyval=(48-height)%2
Call o st /* top border */
xl.='|'
Do y=ymax To ymin By -1
Do s=1 To step
xl.y=xl.y||substr(ol.s.y,xmin,xmax-xmin+1)'|'
End
End
Do y=ymax To ymin By -1
Call o ' 'xl.y
End
Call o sb /* bottom border */
Call lineout oid
Say 'frames are shown in' oid
If debug Then Do
Say 'original area' 1 ml '/' 1 yy
Say 'compute area ' xmi xma '/' ymi yma
Say 'used area ' xlo xhi '/' ylo yhi
End
Do s=1 To step
call lineout oil,s '==>' words(list.s) '==' list.s
End
Return
o: Parse Arg lili
Call lineout oid,lili
Return
set: Parse Arg x,y
a.x.y='*'
Return
neighbors: Procedure Expose a. debug
Parse Arg x,y
neighbors=0
do xa=x-1 to x+1
do ya=y-1 to y+1
If xa<>x | ya<>y then
If a.xa.ya='*' Then
neighbors=neighbors+1
End
End
Return neighbors
store:
/* store current state (a.) in lines m.step.* */
Do y=yma To ymi By -1
ol=''
Do x=xmi To xma
z=a.x.y
ol=ol||z
End
x.step.y=ol
If ol<>'' then Do
ymin=min(ymin,y)
ymax=max(ymax,y)
p=pos('*',ol)
q=length(strip(ol,'T'))
If p>0 Then
xmin=min(xmin,p)
xmax=max(xmax,q)
End
m.step.y=ol
ol.step.y=ol
--If pos('*',ol)>0 Then Do
-- Say '====>' right(step,2) right(y,3) '>'ol'<' xmin xmax
-- Say ' 'copies('1234567890',3)
-- End
End
Return
is_equal:
/* test ist state a.b is equal to state a.a */
Parse Arg a,b
Do y=yy To 1 By -1
If x.b.y<>x.a.y Then
Return 0
End
Return 1
show: Procedure Expose dbg a. yy ml debug
Do y=-5 To 13
ol='>'
Do x=-5 To 13
ol=ol||a.x.y
End
Call debug ol
End
Return
show_neighbors: Procedure Expose a. xmi xma ymi yma dbg debug
Do y=yma To ymi By -1
ol=format(y,3)' '
Do x=xmi To xma
ol=ol||neighbors(x,y)
End
Call debug ol
End
Return
debug:
If debug Then
Return lineout(dbg,arg(1))
Else
Return
|
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #FreeBASIC | FreeBASIC | Dim a As Integer = 1
If a = 1 Then
sub1
ElseIf a = 2 Then
sub2
Else
sub3
End If |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #PicoLisp | PicoLisp | (for L '([] ["ABC"] ["ABC", "DEF"] ["ABC", "DEF", "G", "H"])
(let H (head -1 L)
(prinl
"{"
(glue ", " H)
(and H " and ")
(last L)
"}" ) ) ) |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #PL.2FI | PL/I | *process or(!);
quib: Proc Options(main);
/*********************************************************************
* 06.10.2013 Walter Pachl
*********************************************************************/
put Edit*process or(!);
quib: Proc Options(main);
/*********************************************************************
* 06.10.2013 Walter Pachl
* 07.10.2013 -"- change "Oxford comma" to and
*********************************************************************/
put Edit(quibbling(''))(Skip,a);
put Edit(quibbling('ABC'))(Skip,a);
put Edit(quibbling('ABC DEF'))(Skip,a);
put Edit(quibbling('ABC DEF G H'))(Skip,a);
return;
quibbling: proc(s) Returns(Char(100) Var);
Dcl s Char(*);
Dcl result Char(100) Var Init('');
Dcl word(10) Char(100) Var;
Dcl (wi,p) Bin Fixed(31);
If s='' Then result='';
Else Do;
Do wi=1 By 1 While(s^='');
p=index(s,' ');
if p=0 Then Do;
word(wi)=s;
s='';
End;
Else Do;
word(wi)=left(s,p-1);
s=substr(s,p+1);
End;
end;
wn=wi-1;
result=word(1);
Do i=2 To wn-1;
result=result!!', '!!word(i);
End;
If wn>1 Then
result=result!!' and '!!word(wn);
End;
Return('{'!!result!!'}');
End;
End; |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Scheme | Scheme | (define (main args)
(for-each (lambda (arg) (display arg) (newline)) args)) |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Seed7 | Seed7 | $ include "seed7_05.s7i";
const proc: main is func
local
var integer: i is 0;
begin
writeln("This program is named " <& name(PROGRAM) <& ".");
for i range 1 to length(argv(PROGRAM)) do
writeln("The argument #" <& i <& " is " <& argv(PROGRAM)[i]);
end for;
end func; |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Sidef | Sidef | say ARGV; |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Golfscript | Golfscript | # end of line comment |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Gri | Gri | # this is a comment
show 123 # this too is a comment |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #Oz | Oz | declare
Rules = [rule(c:1 n:[0 1] new:0) %% Lonely
rule(c:1 n:[4 5 6 7 8] new:0) %% Overcrowded
rule(c:1 n:[2 3] new:1) %% Lives
rule(c:0 n:[3] new:1) %% It takes three to give birth!
rule(c:0 n:[0 1 2 4 5 6 7 8] new:0) %% Barren
]
Blinker = ["..."
"###"
"..."]
Toad = ["...."
".###"
"###."
"...."]
Glider = [".#.........."
"..#........."
"###........."
"............"
"............"
"............"
"............"
"............"
"............"
"............"
"............"]
Init = Blinker
MaxGen = 2
%% G(i) -> G(i+1)
fun {Evolve Gi}
fun {Get X#Y}
Row = {CondSelect Gi Y unit}
in
{CondSelect Row X 0} %% cells beyond boundaries are dead (0)
end
fun {GetNeighbors X Y}
{Map [X-1#Y-1 X#Y-1 X+1#Y-1
X-1#Y X+1#Y
X-1#Y+1 X#Y+1 X+1#Y+1]
Get}
end
in
{Record.mapInd Gi
fun {$ Y Row}
{Record.mapInd Row
fun {$ X C}
N = {Sum {GetNeighbors X Y}}
in
for Rule in Rules return:Return do
if C == Rule.c andthen {Member N Rule.n} then
{Return Rule.new}
end
end
end}
end}
end
%% For example: [".#"
%% "#."] -> grid(1:row(1:0 2:1) 2:row(1:1 2:0))
fun {ReadG LinesList}
{List.toTuple grid
{Map LinesList
fun {$ Line}
{List.toTuple row
{Map Line
fun {$ C}
if C == &. then 0
elseif C == &# then 1
end
end}}
end}}
end
%% Inverse to ReadG
fun {ShowG G}
{Map {Record.toList G}
fun {$ Row}
{Map {Record.toList Row}
fun {$ C}
if C == 0 then &.
elseif C == 1 then &#
end
end}
end}
end
%% Helpers
fun {Sum Xs} {FoldL Xs Number.'+' 0} end
fun lazy {Iterate F V} V|{Iterate F {F V}} end
G0 = {ReadG Init}
Gn = {Iterate Evolve G0}
in
for
Gi in Gn
I in 0..MaxGen
do
{System.showInfo "\nGen. "#I}
{ForAll {ShowG Gi} System.showInfo}
end |
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #friendly_interactive_shell | friendly interactive shell | set var 'Hello World'
if test $var = 'Hello World'
echo 'Welcome.'
else if test $var = 'Bye World'
echo 'Bye.'
else
echo 'Huh?'
end |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #PL.2FM | PL/M | 100H:
/* COPY A STRING (MINUS TERMINATOR), RETURNS LENGTH (MINUS TERMINATOR) */
COPY$STR: PROCEDURE(SRC, DST) ADDRESS;
DECLARE (SRC, DST) ADDRESS;
DECLARE (SCH BASED SRC, DCH BASED DST) BYTE;
DECLARE L ADDRESS;
L = 0;
DO WHILE SCH <> '$';
DCH = SCH;
SRC = SRC + 1;
DST = DST + 1;
L = L + 1;
END;
RETURN L;
END COPY$STR;
/* QUIBBLE GIVEN ARRAY OF $-TERMINATED STRINGS, STORE RESULT IN BUFR */
QUIBBLE: PROCEDURE(WORDS, BUFR) ADDRESS;
DECLARE (WORDS, BUFR, ADR) ADDRESS;
DECLARE (WORD BASED WORDS, WPTR) ADDRESS;
DECLARE (WCHAR BASED WPTR, BCHAR BASED BUFR) BYTE;
/* BRACES AND LOWERCASE LETTERS ARE NOT WITHIN PL/M CHARSET */
DECLARE LBRACE LITERALLY '123', RBRACE LITERALLY '125';
DECLARE ANDSTR DATA (32,97,110,100,32,'$');
ADR = BUFR;
BCHAR = LBRACE;
BUFR = BUFR + 1;
DO WHILE WORD <> 0;
BUFR = BUFR + COPY$STR(WORD, BUFR);
WORDS = WORDS + 2;
IF WORD <> 0 THEN
IF WORD(1) <> 0 THEN
BUFR = BUFR + COPY$STR(.', $', BUFR);
ELSE
BUFR = BUFR + COPY$STR(.ANDSTR, BUFR);
END;
BCHAR = RBRACE;
BUFR = BUFR + 1;
BCHAR = '$';
RETURN ADR;
END QUIBBLE;
/* --- CP/M OUTPUT AND TESTING --- */
BDOS: PROCEDURE(FUNC, ARG); /* MAKE CP/M SYSTEM CALL */
DECLARE FUNC BYTE, ARG ADDRESS;
GO TO 5;
END BDOS;
DECLARE BDOS$EXIT LITERALLY '0', /* EXIT TO CP/M */
BDOS$PUTS LITERALLY '9'; /* PRINT STRING */
PUTS: PROCEDURE(S);
DECLARE S ADDRESS;
CALL BDOS(BDOS$PUTS, S);
CALL BDOS(BDOS$PUTS, .(13,10,'$'));
END PUTS;
/* ARRAY WITH INITIALLY NO CONTENTS */
DECLARE ARR (5) ADDRESS INITIAL (0,0,0,0,0);
CALL PUTS(QUIBBLE(.ARR, .MEMORY)); /* NO STRINGS */
ARR(0) = .'ABC$';
CALL PUTS(QUIBBLE(.ARR, .MEMORY)); /* ABC */
ARR(1) = .'DEF$';
CALL PUTS(QUIBBLE(.ARR, .MEMORY)); /* ABC AND DEF */
ARR(2) = .'G$';
ARR(3) = .'H$';
CALL PUTS(QUIBBLE(.ARR, .MEMORY)); /* ABC, DEF, G AND H */
CALL BDOS(BDOS$EXIT, 0);
EOF |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #Plain_English | Plain English | To quibble four words:
Add "ABC" to some string things.
Add "DEF" to the string things.
Add "G" to the string things.
Add "H" to the string things.
Quibble the string things.
To quibble one word:
Add "ABC" to some string things.
Quibble the string things.
To quibble some string things:
Quibble the string things giving a string.
Destroy the string things.
Write the string on the console.
To quibble some string things giving a string:
Append "{" to the string.
Put the string things' count into a count.
If the count is 0, append "}" to the string; exit.
Get a string thing from the string things.
If the count is 1, append the string thing's string then "}" to the string; exit.
Loop.
If a counter is past the count minus 2, append the string thing's string then " and " then the string thing's next's string then "}" to the string; exit.
Append the string thing's string then ", " to the string.
Put the string thing's next into the string thing.
Repeat.
To quibble two words:
Add "ABC" to some string things.
Add "DEF" to the string things.
Quibble the string things.
To quibble zero words:
Quibble some string things.
To run:
Start up.
Quibble zero words.
Quibble one word.
Quibble two words.
Quibble four words.
Wait for the escape key.
Shut down. |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Slate | Slate | StartupArguments do: [| :arg | inform: arg] |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Smalltalk | Smalltalk | (1 to: Smalltalk getArgc) do: [ :i |
(Smalltalk getArgv: i) displayNl
] |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Standard_ML | Standard ML | print ("This program is named " ^ CommandLine.name () ^ ".\n");
val args = CommandLine.arguments ();
Array.appi
(fn (i, x) => print ("the argument #" ^ Int.toString (i+1) ^ " is " ^ x ^ "\n"))
(Array.fromList args); |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Groovy | Groovy | 100 REM Standard BASIC comments begin with "REM" (remark) and extend to the end of the line
110 PRINT "this is code": REM comment after statement |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #GW-BASIC | GW-BASIC | 100 REM Standard BASIC comments begin with "REM" (remark) and extend to the end of the line
110 PRINT "this is code": REM comment after statement |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #PARI.2FGP | PARI/GP | step(M)={
my(N=M,W=matsize(M)[1],L=#M,t);
for(l=1,W,for(w=1,L,
t=sum(i=l-1,l+1,sum(j=w-1,w+1,if(i<1||j<1||i>W||j>L,0,M[i,j])));
N[l,w]=(t==3||(M[l,w]&&t==4))
));
N
};
M=[0,1,0;0,1,0;0,1,0];
for(i=1,3,print(M);M=step(M)) |
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #Futhark | Futhark |
if <condition> then <truebranch> else <falsebranch>
|
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #PowerShell | PowerShell |
function Out-Quibble
{
[OutputType([string])]
Param
(
# Zero or more strings.
[Parameter(Mandatory=$false, Position=0)]
[AllowEmptyString()]
[string[]]
$Text = ""
)
# If not null or empty...
if ($Text)
{
# Remove empty strings from the array.
$text = "$Text".Split(" ", [StringSplitOptions]::RemoveEmptyEntries)
}
else
{
return "{}"
}
# Build a format string.
$outStr = ""
for ($i = 0; $i -lt $text.Count; $i++)
{
$outStr += "{$i}, "
}
$outStr = $outStr.TrimEnd(", ")
# If more than one word, insert " and" at last comma position.
if ($text.Count -gt 1)
{
$cIndex = $outStr.LastIndexOf(",")
$outStr = $outStr.Remove($cIndex,1).Insert($cIndex," and")
}
# Output the formatted string.
"{" + $outStr -f $text + "}"
}
|
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Swift | Swift | let args = Process.arguments
println("This program is named \(args[0]).")
println("There are \(args.count-1) arguments.")
for i in 1..<args.count {
println("the argument #\(i) is \(args[i])")
} |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Tailspin | Tailspin |
$ARGS -> !OUT::write
|
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Tcl | Tcl | if { $argc > 1 } {
puts [lindex $argv 1]
} |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Haskell | Haskell | i code = True -- I am a comment.
{- I am also
a comment. {-comments can be nested-}
let u x = x x (this code not compiled)
Are you? -}
-- |This is a Haddock documentation comment for the following code
i code = True
-- ^This is a Haddock documentation comment for the preceding code
{-|
This is a Haddock documentation block comment
-}
i code = True |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Haxe | Haxe | // Single line commment.
/*
Multiple
line
comment.
*/ |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #Pascal | Pascal | program Gol;
// Game of life
{$IFDEF FPC}
//save as gol.pp/gol.pas
{$Mode delphi}
{$ELSE}
//for Delphi save as gol.dpr
{$Apptype Console}
{$ENDIF}
uses
crt;
const
colMax = 76;
rowMax = 22;
dr = colMax+2; // element count of one row
cDelay = 20; // delay in ms
(*
expand field by one row/column before and after
for easier access no special treatment of torus
*)
type
tFldElem = byte;//0..1
tpFldElem = ^tFldElem;
tRow = array[0..colMax+1] of tFldElem;
tpRow = ^tRow;
tBoard = array[0..rowMax+1] of tRow;
tpBoard = ^tBoard;
tpBoards = array[0..1] of tpBoard;
type
tIntArr = array[0..2*dr+2] of tFldElem;
tpIntArr = ^tIntArr;
var
aBoard,bBoard : tBoard;
pBoards :tpBoards;
gblActBoard : byte;
gblUseTorus :boolean;
gblGenCnt : integer;
procedure PrintGen;
const
cChar: array[0..1] of char = (' ','#');
var
p0 : tpIntArr;
col,row: integer;
s : string[colMax];
begin
setlength(s,colmax);
gotoxy(1,1);
writeln(gblGenCnt:10);
For row := 1 to rowMax do
begin
p0 := @pBoards[gblActBoard]^[row,0];;
For col := 1 to colMax do
s[col] := cChar[p0[col]];
writeln(s);
end;
delay(cDelay);
end;
procedure Init0(useTorus:boolean);
begin
gblUseTorus := useTorus;
gblGenCnt := 0;
fillchar(aBoard,SizeOf(aBoard),#0);
pBoards[0] := @aBoard;
pBoards[1] := @bBoard;
gblActBoard := 0;
clrscr;
end;
procedure InitRandom(useTorus:boolean);
var
col,row : integer;
begin
Init0(useTorus);
For row := 1 to rowMax do
For col := 1 to colMax do
aBoard[row,col]:= tFldElem(random>0.9);
end;
procedure InitBlinker(useTorus:boolean);
var
col,row : integer;
begin
Init0(useTorus);
For col := 1 to colMax do
begin
IF (col+2) mod 4 = 0 then
begin
For row := 1 to rowmax do
IF row mod 4 <> 0 then
aBoard[row,col]:= 1;
end;
end;
end;
procedure Torus;
var
p0 : tpIntArr;
row: integer;
begin
//copy column 1-> colMax+1 and colMax-> 0
p0 := @pBoards[gblActBoard]^[1,0];
For row := 1 to rowMax do
begin
p0^[0] := p0^[colMax];
p0^[colmax+1] := p0^[1];
//next row
p0 := Pointer(PtrUint(p0)+SizeOf(tRow));
end;
//copy row 1-> rowMax+1
move(pBoards[gblActBoard]^[1,0],pBoards[gblActBoard]^[rowMax+1,0],sizeof(trow));
//copy row rowMax-> 0
move(pBoards[gblActBoard]^[rowMax,0],pBoards[gblActBoard]^[0,0],sizeof(trow));
end;
function Survive(p: tpIntArr):tFldElem;
//p points to actual_board [row-1,col-1]
//calculates the sum of alive around [row,col] aka p^[dr+1]
//really fast using fpc 2.6.4 no element on stack
const
cSurvives : array[boolean,0..8] of byte =
//0,1,2,3,4,5,6,7,8 sum of alive neighbours
((0,0,0,1,0,0,0,0,0), {alive =false 1->born}
(0,0,1,1,0,0,0,0,0)); {alive =true 0->die }
var
sum : integer;
begin
// row above
// sum := byte(aBoard[row-1,col-1])+byte(aBoard[row-1,col])+byte(aBoard[row-1,col+1]);
sum := integer(p^[ 0])+integer(p^[ 1])+integer(p^[ 2]);
sum := sum+integer(p^[ dr+0]) +integer(p^[ dr+2]);
sum := sum+integer(p^[2*dr+0])+integer(p^[2*dr+1])+integer(p^[2*dr+2]);
survive := cSurvives[boolean(p^[dr+1]),sum];
end;
procedure NextGen;
var
p0,p1 : tpFldElem;
row: NativeInt;
col :NativeInt;
begin
if gblUseTorus then
Torus;
p1 := @pBoards[1-gblActBoard]^[1,1];
//One row above and one column before because of survive
p0 := @pBoards[ gblActBoard]^[0,0];
For row := rowMax-1 downto 0 do
begin
For col := colMax-1 downto 0 do
begin
p1^ := survive(tpIntArr(p0));
inc(p0);
inc(p1);
end;
// jump over the borders
inc(p1,2);
inc(p0,2);
end;
//aBoard := bBoard;
gblActBoard :=1-gblActBoard;
inc(gblGenCnt);
end;
begin
InitBlinker(false);
repeat
PrintGen;
NextGen;
until keypressed;
PrintGen;
end.
|
http://rosettacode.org/wiki/Conditional_structures | Conditional structures | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
List the conditional structures offered by a programming language. See Wikipedia: conditionals for descriptions.
Common conditional structures include if-then-else and switch.
Less common are arithmetic if, ternary operator and Hash-based conditionals.
Arithmetic if allows tight control over computed gotos, which optimizers have a hard time to figure out.
| #GAP | GAP | if <condition> then
<statements>
elif <condition> then
<statements>
else
<statements>
fi; |
http://rosettacode.org/wiki/Comma_quibbling | Comma quibbling | Comma quibbling is a task originally set by Eric Lippert in his blog.
Task
Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
An input of no words produces the output string of just the two brace characters "{}".
An input of just one word, e.g. ["ABC"], produces the output string of the word inside the two braces, e.g. "{ABC}".
An input of two words, e.g. ["ABC", "DEF"], produces the output string of the two words inside the two braces with the words separated by the string " and ", e.g. "{ABC and DEF}".
An input of three or more words, e.g. ["ABC", "DEF", "G", "H"], produces the output string of all but the last word separated by ", " with the last word separated by " and " and all within braces; e.g. "{ABC, DEF, G and H}".
Test your function with the following series of inputs showing your output here on this page:
[] # (No input words).
["ABC"]
["ABC", "DEF"]
["ABC", "DEF", "G", "H"]
Note: Assume words are non-empty strings of uppercase characters for this task.
| #Prolog | Prolog | words_series(Words, Bracketed) :-
words_serialized(Words, Serialized),
atomics_to_string(["{",Serialized,"}"], Bracketed).
words_serialized([], "").
words_serialized([Word], Word) :- !.
words_serialized(Words, Serialized) :-
append(Rest, [Last], Words), %% Splits the list of *Words* into the *Last* word and the *Rest*
atomics_to_string(Rest, ", ", WithCommas),
atomics_to_string([WithCommas, " and ", Last], Serialized).
test :-
forall( member(Words, [[], ["ABC"], ["ABC", "DEF"], ["ABC", "DEF", "G", "H"]]),
( words_series(Words, Series),
format('~w ~15|=> ~w~n', [Words, Series]))
). |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #Toka | Toka | [ arglist array.get type cr ] is show-arg
[ dup . char: = emit space ] is #=
1 #args [ i #= show-arg ] countedLoop |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #TXR | TXR | @(next :args)
@(collect)
@arg
@(end)
@(output)
My args are: {@(rep)@arg, @(last)@arg@(end)}
@(end) |
http://rosettacode.org/wiki/Command-line_arguments | Command-line arguments | Command-line arguments is part of Short Circuit's Console Program Basics selection.
Scripted main
See also Program name.
For parsing command line arguments intelligently, see Parsing command-line arguments.
Example command line:
myprogram -c "alpha beta" -h "gamma"
| #UNIX_Shell | UNIX Shell | WHOLELIST="$@" |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #HicEst | HicEst | ! a comment starts with a "!" and ends at the end of the line |
http://rosettacode.org/wiki/Comments | Comments | Task
Show all ways to include text in a language source file
that's completely ignored by the compiler or interpreter.
Related tasks
Documentation
Here_document
See also
Wikipedia
xkcd (Humor: hand gesture denoting // for "commenting out" people.)
| #Hope | Hope | ! All Hope comments begin with "!" and extend to the end of the line |
http://rosettacode.org/wiki/Conway%27s_Game_of_Life | Conway's Game of Life | The Game of Life is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the best-known example of a cellular automaton.
Conway's game of life is described here:
A cell C is represented by a 1 when alive, or 0 when dead, in an m-by-m (or m×m) square array of cells.
We calculate N - the sum of live cells in C's eight-location neighbourhood, then cell C is alive or dead in the next generation based on the following table:
C N new C
1 0,1 -> 0 # Lonely
1 4,5,6,7,8 -> 0 # Overcrowded
1 2,3 -> 1 # Lives
0 3 -> 1 # It takes three to give birth!
0 0,1,2,4,5,6,7,8 -> 0 # Barren
Assume cells beyond the boundary are always dead.
The "game" is actually a zero-player game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
Task
Although you should test your implementation on more complex examples such as the glider in a larger universe, show the action of the blinker (three adjoining cells in a row all alive), over three generations, in a 3 by 3 grid.
References
Its creator John Conway, explains the game of life. Video from numberphile on youtube.
John Conway Inventing Game of Life - Numberphile video.
Related task
Langton's ant - another well known cellular automaton.
| #Perl | Perl | life.pl numrows numcols numiterations
life.pl 5 10 15
|
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