[Bug tree-optimization/65752] Too strong optimizations int -> pointer casts

["gil.hur at sf dot snu.ac.kr" <gcc-bugzilla at gcc dot gnu dot org>]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=65752

--- Comment #21 from Chung-Kil Hur <gil.hur at sf dot snu.ac.kr> ---
(In reply to Marek Polacek from comment #20)
> (In reply to Chung-Kil Hur from comment #19)
> > (In reply to rguenther@suse.de from comment #18)
> > > On Tue, 19 May 2015, gil.hur at sf dot snu.ac.kr wrote:
> > > 
> > > > https://gcc.gnu.org/bugzilla/show_bug.cgi?id=65752
> > > > 
> > > > --- Comment #17 from Chung-Kil Hur <gil.hur at sf dot snu.ac.kr> ---
> > > > Hi Richard,
> > > > 
> > > > I modified the example further.
> > > > 
> > > > #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;
> > > >   /* The following "if" statement is never executed because j == xp */
> > > >   if (j != xp) { 
> > > >     printf("hello\n");
> > > >     j = xp; 
> > > >   }
> > > 
> > > Here j is always xp and thus ...
> > > 
> > 
> > Why is "j" always "xp"?
> > Since "hello" is not printed, "j = xp;" is not executed.
> 
> Because that "if (j != xp)" guarantees it.

OK. here is another modification.

#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*)j = 15;

  /* The following "if" statement is never executed because j == xp */
  if (j != xp) { 
    printf("hello\n");
    j = xp; 
  }

  *(int*)((xp+i)-j) = 15;

  printf("%d\n", x);
}

This program just prints "0".

So we know that "*(int*)j = 15;" is not executed and thus "j == xp" is not
true.

Then, can the following statement change "j" even if the printf is not
executed?

if (j != xp) {
   printf("hello\n");
   j = xp;
}

If not, how can "j == xp" suddenly hold?