wrong code at -O1/-O2 on x86_64-linux-gnu

[tangyixuan01]

Compiler explorer: https://godbolt.org/z/PP5eaEccb

Clang-18 produces the wrong code when compiling the following code with "-O1" or "-O2", while outputting the correct result with the optimization flag "-O3".

int printf(const char *, ...);
long a;
int b, c, e, g, i;
long *d, *h;
char f = 6;
int main() {
  long j;
  c = 0;
  for (; c != 7; ++c) {
    long k;
    long **l = &d;
    for (; f + i; i++)
      h = &k;
    g = h != (*l = &j);
    int *m = &b;
    *m = g;
    for (; e; a = a + 1)
      ;
  }
  printf("%d\n", b);
}

clang -O2 s.c; ./a.out
0

clang -O3 s.c; ./a.out
1

gcc -O2 s.c; ./a.out
1

[dtcxzyw]

gcc -O0 s.c; ./a.out hangs. Is there an infinite loop?

[EugeneZelenko]

gcc -O0 s.c; ./a.out hangs. Is there an infinite loop?

Third loops seems to be such.

[dtcxzyw]

I believe this code has UB as for (; e; a = a + 1); is a non-trivial infinite loop.

[dtcxzyw]

I guess you are using creduce. If you think the original case doesn't contain UB, can you share the code?

[tangyixuan01]

I guess you are using creduce. If you think the original case doesn't contain UB, can you share the code?

Thanks for your reply!
Yes, I use creduce for the code reduction. I reduce it again, and get the same results after the reduction.
Compiler explorer: https://godbolt.org/z/jjxx9srM3

Please consider the following code:

int printf(const char *, ...);
int b, c, g, i;
long *d, *h;
int main() {
  long j;
  c = 0; 
  for (; c <1; ++c) {
    long k;
    long **l = &d;
    for (; i<1; i++)
      h = &k;
    g = h != (*l = &j);
    int *m = &b;
    *m = g;
  }
  printf("%d\n", b);
}

> clang -O2 s.c; ./a.out
0

> clang -O0 s.c; a.out
1