On 17/08/16 02:21, Toshi Morita wrote:
I was involved in a discussion over the semantics of "const" in C, and the
following code was posted:
#include <stdio.h>
int foo = 0;
const int *pfoo = &foo;
void bar (void)
{
foo +=3D;
I assume that's a typo?
}
int main(void)
{
int a, b;
a = *pfoo;
bar();
b = *pfoo;
printf("a: %d, b: %d\n", a, b);
}
This code when compiled with gcc 4.8.2 using the optimization option -O3
produces:
a: 0, b: 1
So it appears even though pfoo is a const int *, the value *pfoo is read twice.
Would it be valid for the code to print a:0, b: 0?
If so, is this a missed optimization opportunity?
No, it would not be valid. Declaring pfoo as a "const int*" tells the
compiler "I will not change anything via this pointer - and you can
optimise based on that promise". It does /not/ tell the compiler "the
thing that this points to will not change".
So the compiler is correct in reading *pfoo twice.
