https://gcc.gnu.org/bugzilla/show_bug.cgi?id=78420
--- Comment #3 from Tomasz Kamiński <tomaszkam at gmail dot com> --- > I don't see this as prohibiting the transformation. The standard seems to be > saying that they might or might not compare as equal, which presumably > depends on how variables are laid out in memory. The optimization seems > wrong to me. But, the code snippet is not using == on this pointer, but std::less specialization that is required to form total order, but in the attached scenario as not because for p and b nor of following hold: 1) r(p, b) 2) r(b, p) 3) b is same as p (!r(p, b) && !r(b, p)) Which proves that r is not an total order. So the guarantee on the less<T*> is no longer provided.
