https://gcc.gnu.org/bugzilla/show_bug.cgi?id=50417
--- Comment #9 from Oleg Endo <olegendo at gcc dot gnu.org> ---
(In reply to rguent...@suse.de from comment #8)
>
> But there is no good reasoning that can be applied that it is a valid
> transform. What does clang do when s is void * and you cast that to
> uint32_t *? Note that memcpy prototype takes a void * argument.
Does it matter whether there's a void* in between or not?
If we do a *(int32_T*)(void*)(int32_t*)p on a strict alignment target, it will
result in a 32 bit mem access.
GCC knows memcpy and its semantics well enough so propagating the target's
alignment rules into memcpy is most likely going to be a safe and sane thing to
do. So at least cases like
void test (const int* a, int* b)
{
std::memcpy (b, a, 4);
}
should work. Shouldn't it?
