https://gcc.gnu.org/bugzilla/show_bug.cgi?id=65752
--- Comment #24 from Chung-Kil Hur <gil.hur at sf dot snu.ac.kr> ---
(In reply to schwab from comment #23)
> "gil.hur at sf dot snu.ac.kr" <gcc-bugzi...@gcc.gnu.org> writes:
>
> > Since "hello" is not printed, I think the if-statement is the same as no-op.
> > Thus, removing the if-statement should not change the behavior of the
> > program
> > according to ISO C11.
>
> Unless you are invoking undefined behaviour.
>
> Andreas.
==============================
#include <stdio.h>
int main() {
int x = 0;
uintptr_t xp = (uintptr_t) &x;
uintptr_t i, j;
for (i = 0; i < xp; i++) { }
j = i;
*(int*)((xp+i)-j) = 15;
printf("%d\n", x);
}
=============================
This prints "15".
And I do not think there is any UB.
Please correct me if I am wrong.
Then, I add the if-statement.
==============================
#include <stdio.h>
int main() {
int x = 0;
uintptr_t xp = (uintptr_t) &x;
uintptr_t i, j;
for (i = 0; i < xp; i++) { }
j = i;
/****** begin *******/
if (j != xp) {
printf("hello\n");
j = xp;
}
/****** end *********/
*(int*)((xp+i)-j) = 15;
printf("%d\n", x);
}
=============================
This prints "0" without printing "hello".
Thus, this raises no UB unless "j != xp" raises UB.
If you think "j != xp" raises UB, please explain why and give some reference.
Otherwise, I think it is a bug of GCC.
