problem with bitwise OR I have this code, in which I try to "glue" four bytes in order to construct an unsigned int. I shift left the bytes and "add" them using the OR operator.

unsigned long number1;   
unsigned long number2;
unsigned long number3=0;

byte byte0 = 1;
byte byte1 = 40;
byte byte2 = 107;
byte byte3 = 238;

void setup(){
  Serial.begin(9600);
}

void loop(){
  number1 = byte1 &lt;&lt;8 | byte0 ;  //(=10.241 O.K.)

  //(should be 4.000.000.001, it produces 10.241)
  number2 = byte3 &lt;&lt;24 | byte2 &lt;&lt;16 | byte1 &lt;&lt;8 | byte0;  

  //(should be 4.000.000.001, it produces 4.016.777.217)
  number3 = number3 | byte3;
  number3 = number3&lt;&lt;8;
  number3 = number3 | byte2;
  number3 = number3&lt;&lt;8;
  number3 = number3 | byte1;
  number3 = number3&lt;&lt;8;
  number3 = number3 | byte0;  

  Serial.print("number1 = ");Serial.println(number1);  
  Serial.print("number2 = ");Serial.println(number2);   
  Serial.print("number3 = ");Serial.println(number3);       
  Serial.println(" ");  
}


The problem is that the OR operator applies only to byte1 &lt;&lt;8 | byte0 and ignores the rest. That is number1 and number2 appear to have the same value (10241). 

I also tried "adding" them up separately, as shown with number3, but it also produces a wrong result as well.

I tried the same code in python, and it works fine.

byte3, byte2, byte1, byte0 = 238, 107, 40, 1
print byte1 &lt;&lt;8 | byte0
print byte3 &lt;&lt;24 | byte2 &lt;&lt;16 | byte1 &lt;&lt;8 | byte0

number3 = 0;
number3 = number3 | byte3;
number3 = number3&lt;&lt;8;
number3 = number3 | byte2;
number3 = number3&lt;&lt;8;
number3 = number3 | byte1;
number3 = number3&lt;&lt;8;
number3 = number3 | byte0;
print number3


So, the question is:

Why both number2 and number3 produce a false result?
Correction from another answer for number2:

number2 = (unsigned long)byte3 &lt;&lt;24 | (unsigned long)byte2 &lt;&lt;16 | byte1 &lt;&lt;8 | byte0; // = 4.000.000.001


And in the first round of loop number3 is correct, all others are wrong as number3 holds a result from previous loop.

  number3 = byte3; // all magic lays here
  number3 &lt;&lt;= 8;
  number3 |= byte2;
  number3 &lt;&lt;= 8;
  number3 |= byte1;
  number3 &lt;&lt;= 8;
  number3 |= byte0;