https://gcc.gnu.org/bugzilla/show_bug.cgi?id=77686
--- Comment #17 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to Richard Biener from comment #14)
> Just to explain a little bit. Consider
>
> void foo (void)
> {
> union { float f; char c[4]; } u, v;
> *(int *)&u = 0;
> v = u;
> return *(int *)&v;
> }
>
> then C11 6.5/6,7 make it clear that it is not valid to use 'u' (of union
> type)
> to access the object which now has effective type 'int' (in fact that store
> alone, given strict reading of C11 6.5/6 is invalid as u has a declared
> type).
> The union does not have 'int' amongst its members. Putting 'char' or an
> array of 'char' into the union does not make the access fall under 6.5/7
> as 'a character type' only follows 'an aggregate or union type that includes
> one
> of the _AFOREMENTIONED_ types...' (emphasis mine).
But would *(int*)u.c be OK, because that is a character type? Given that C++
does define what the assignment means for the union, we just need to make sure
we do *(int*)u.c not *(int*)&u or is that still not enough?
If not, what does the last bullet of 6.5/7 allow?
