Re: Regarding assignment to struct

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];

So, what you are basing your "UB" claim on is not clear to me.
 

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