Re: Regarding assignment to struct

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);
So, what you are basing your "UB" claim on is not clear to me.
--
Best regards,
Andrey