Why this loop is not optimised out?

[Ian Lance Taylor]

jgojgo <chairmanguo@gmail.com> writes:

> I've got a very simple c program which copies all elements from array A to
> back to array A. For example,
>
> int *A;
> A = (int*)malloc(sizeof(int)*SIZE);
> for( i = 0; i < SIZE; i++) {
>   A[i] = A[i];
> }
> I was expecting this to be optimised out by the compiler and eventually
> turned into a noop. However, by measuring the runtime of this loop and
> looking at the assembly code, it seems that the element is indeed loaded
> from memory into register and then stored back to the same memory location.
> I have -O3 enabled. Can anyone explain to me why the c compiler does not
> optimise it? Or am I missing something here?

You neglected to mention which version of the compiler you tested.

I turned your fragment into a real program:

extern void *malloc (long unsigned int);
#define SIZE 100
void f()
{
  int *A;
  int i;
  A = (int*)malloc(sizeof(int)*SIZE);
  for( i = 0; i < SIZE; i++) {
    A[i] = A[i];
  }
}

I compiled it with -O2 using current mainline sources.  The entire loop
was eliminated and the function turned into a no-op.

Ian