Tim Rentsch <tr.17687@z991.linuxsc.com> writes:
>Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:>
>Andrey Tarasevich <noone@noone.net> writes:>
[...]
>#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.
I believe it does. [...]
If you look again carefully, I expect you will reach a
different conclusion.
Please explain. Or don't.
>
I explained, in text that you snipped, why I think it has UB.
If there's a flaw in my reasoning, I invite you to point it out
without being mysterious.
I wasn't thinking about your reasoning or what flaw there may have
been in it. My intention was to suggest a course of action that
would be helpful rather than to point out any sort of mistake. But
since you ask, I think the mistake you made was jumping to a
conclusion without looking carefully enough at the problem.
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