text: C array subscripts can be understood via pointer arithmetic.
  #include <stdio.h> int main(void) { int vals[10]; printf("%d\n", vals[3]); return 0; }
  What does
  vals[3] mean here? It means “take the pointer
  vals, shift it up to the fourth element in the array, and dereference that pointer”. We can write this out as
  *(vals + 3).
  This is in fact the definition of the array subscript operator in C:
  a[b] means
  *(a+b).
  The Standard says
  For addition, either both operands shall have arithmetic type, or one operand shall be a pointer to an object type and the other shall have integer type. (Incrementing is equivalent to adding 1.)
  Above, we are in the latter case: “one operand shall be a pointer to an object type and the other shall have integer type”.
  vals has type
  int*, and
  3 has integer type.
  When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand.
  So the expression
  vals+3 has type
  int*.
  the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression.
  That is, if we consider that
  vals actually points to some
  arr[n], then
  vals+3 points to
  arr[n+3].
  Note that this definition is commutative:
  3+vals also points to
  arr[n+3].
  This has the funny consequence that
  a[b] is actually the same as
  b[a]!
  I wrote this because I felt like it. This post is my own, and not associated with my employer.Jim. Public speaking. Friends. Vidrio.