https://gcc.gnu.org/bugzilla/show_bug.cgi?id=82898
--- Comment #4 from rguenther at suse dot de <rguenther at suse dot de> --- On Wed, 8 Nov 2017, antoshkka at gmail dot com wrote: > https://gcc.gnu.org/bugzilla/show_bug.cgi?id=82898 > > --- Comment #3 from Antony Polukhin <antoshkka at gmail dot com> --- > > At least in the gcc model, the type of a pointer is meaningless > > as long as you do not dereference it using that type<...> > > OK, does dereferencing the pointers change anything? > > void foo(int* i, const float* f) { > i[0] = f[0]; // Does it triggers the aliasing logic? for GCC it says that at *f the dynamic type at function entry was float and after the assignment that of *i is now int. It doesn't say that f != i. You'd have to do global_i = *i; global_f = *f; which would be only valid if i != f (and thus they are non-overlapping because alignof of both is the same as the size) > __builtin_memmove(i, f, 1024*1024); > } > > Code from above still uses memmove, instead of memcpy. Yes. For bigger sizes than alignof I'm sure it doesn't work. For __builtin_memmove (i, f, 4); it doesn't work because we don't derive context sensitive points-to (and I have a hard time thinking of an efficient and effective way of doing that). For the small sizes we instead optimize the memmove as a read from *f and then a store to *i.
