https://gcc.gnu.org/bugzilla/show_bug.cgi?id=109680
--- Comment #9 from CVS Commits <cvs-commit at gcc dot gnu.org> --- The trunk branch has been updated by Marek Polacek <mpola...@gcc.gnu.org>: https://gcc.gnu.org/g:4c2ffb02fd4104d77c5d907662f04434dc4c3fe8 commit r14-669-g4c2ffb02fd4104d77c5d907662f04434dc4c3fe8 Author: Marek Polacek <pola...@redhat.com> Date: Tue May 2 17:36:00 2023 -0400 c++: wrong std::is_convertible with cv-qual fn [PR109680] This PR points out that std::is_convertible has given the wrong answer in static_assert (!std::is_convertible_v <int () const, int (*) ()>, ""); since r13-2822 implemented __is_{,nothrow_}convertible. std::is_convertible uses the imaginary To test() { return std::declval<From>(); } to do its job. Here, From is 'int () const'. std::declval is defined as: template<class T> typename std::add_rvalue_reference<T>::type declval() noexcept; std::add_rvalue_reference is defined as "If T is a function type that has no cv- or ref- qualifier or an object type, provides a member typedef type which is T&&, otherwise type is T." In our case, T is cv-qualified, so the result is T, so we end up with int () const declval() noexcept; which is invalid. In other words, this is pretty much like: using T = int () const; T fn1(); // bad, fn returning a fn T& fn2(); // bad, cannot declare reference to qualified function type T* fn3(); // bad, cannot declare pointer to qualified function type using U = int (); U fn4(); // bad, fn returning a fn U& fn5(); // OK U* fn6(); // OK I think is_convertible_helper needs to simulate std::declval better. To that end, I'm introducing build_trait_object, to be used where a declval is needed. PR c++/109680 gcc/cp/ChangeLog: * method.cc (build_trait_object): New. (assignable_expr): Use it. (ref_xes_from_temporary): Likewise. (is_convertible_helper): Likewise. Check FUNC_OR_METHOD_TYPE_P. gcc/testsuite/ChangeLog: * g++.dg/ext/is_convertible6.C: New test.
