> > For the purposes of these operators, a pointer to an object that is not an e...

tzs · on July 2, 2018

Then let's make them real arrays:

  #include <stdio.h>

  #define N 8

  int main(void) {
    int a[N], b[N];
    int *p = &a[0];
    int *q = &b[N];
    printf("%p %p %d\n", (void *)p, (void *)q, p == q);
    return 0;
  }

Results:

  $ gcc -std=c11 -O1 ar.c; ./a.out
  0x7ffd32d048c0 0x7ffd32d048c0 0
  $ gcc -std=c11 ar.c; ./a.out
  0x7ffe3f8ccd60 0x7ffe3f8ccd60 1

smolsky · on July 3, 2018

BTW, things work correctly when the objects live inside a real array:

  #include <stdio.h>

  int main(void) {
    int arr[16];
    int *p = &arr[0] + 1;
    int *q = &arr[1];
    printf("pointers: %p %p %d\n", p, q, p == q);
    return 0;
  }

The result is perfectly correct and predictable, as the Standard guarantees pointer comparisons in this case:

  pointers: 0x71b35ac790d4 0x71b35ac790d4 1

The original code in the article relies on UB, I think, so all bets are off.