https://gcc.gnu.org/bugzilla/show_bug.cgi?id=109680
--- Comment #7 from Marek Polacek <mpolacek at gcc dot gnu.org> ---
Ah, I think I see what's going on here. Once again, the problem is that this
assert no longer passes:
#include <utility>
static_assert (!std::is_convertible_v <int () const, int (*) ()>, "");
std::is_convertible does
To test() { return std::declval<From>(); }
here, From is 'int () const'.
std::declval is defined as:
template<class T>
typename std::add_rvalue_reference<T>::type declval() noexcept;
Now, 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:
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
So the check we're looking for is probably
if (TREE_CODE (type) == FUNCTION_TYPE
&& (type_memfn_quals (type) != TYPE_UNQUALIFIED
|| type_memfn_rqual (type) != REF_QUAL_NONE))
but I think it should be put wherever we simulate declval().
