Re: Regarding assignment to struct

On Mon 5/5/2025 2:01 AM, Michael S wrote:

On Mon, 5 May 2025 01:29:47 -0700
Andrey Tarasevich <noone@noone.net> wrote:
 

On Mon 5/5/2025 1:12 AM, Michael S wrote:

>
According to my understanding, you are wrong.
Taking pointer of non-lvalue is UB, so anything compiler does is
conforming.
  

>
Er... What? What specifically do you mean by "taking pointers"?
>
The whole functionality of `[]` operator in C is based on pointers.
This expression
>
    (a = b).a[5]
>

 

is already doing your "taking pointers of non-lvalue" (if I
understood you correctly) as part of array-to-pointer conversion. And
no, it is not UB.
>
This is not UB either
>
    struct S foo(void) { return (struct S) { 1, 2, 3 }; }
    ...
    int *p;
    p = &foo().a[2], printf("%d\n", *p);
>

  That is not UB:
int a5 = (a = b).a[5];
 That is UB:
int* pa5 = &(a = b).a[5];

No, it isn't.

If you read the post of Keith Thompson and it is still not clears to
you then I can not help.

The only valid "UB" claim in Keith's post is my printing the value of `pc` pointer, which by that time happens to point nowhere, since the lifetime of the temporary is over. (And, of course, lack of conversion to `void *` is an issue).
As for the expressions like
   &(a = b).a[5];
and
   &foo().a[2]
- these by themselves are are perfectly valid. There's no UB in these expressions. (And this is not a debate.)
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.
--
Best regards,
Andrey