Re: Regarding assignment to struct

On Mon 5/5/2025 10:20 AM, Michael S wrote:

>
Here's a version of the same code that corrects the above distracting
issues
>
    #include <stdio.h>
>
    struct S { int a[10]; };
>
    int main()
    {
      struct S a, b = { 0 };
      int *pa, *pb, *pc;
>
      pa = &a.a[5],
      pb = &b.a[5],
      pc = &(a = b).a[5],
      printf("%p %p %p\n", (void *) pa, (void *) pb, (void *) pc);
    }
>
This version has no UB.
>

 It's only not UB in the nazal demons sense.
It's UB in a sense that we can't predict values of expressions
like (pa==pc) and (pb==pc). I.e. pc is completely useless. In my book
it is form of UB.

Whether we can or cannot predict the values of `(pa==pc)` and `(pb==pc)` has very little impact on the usability of such expressions. The practical usability of such expressions is very high without relying on `(pa==pc)` and `(pb==pc)`.
--
Best regards,
Andrey