No member named value in std::is_convertible when using clang - gcc

Question

In a very simple situation with a constrained constructor, testing for convertibility of the argument, an error is produced in clang, but not in g++:
#include <type_traits>
template <class T, class U>
constexpr bool Convertible = std::is_convertible<T,U>::value && std::is_convertible<U,T>::value;
template <class T>
struct A
{
template <class S, class = std::enable_if_t<Convertible<S,T>> >
A(S const&) {}
};
int main()
{
A<double> s = 1.0;
}
Maybe this issue is related to Is clang's c++11 support reliable?
The error clang gives, reads:
error: no member named 'value' in 'std::is_convertible<double, A<double> >'
constexpr bool Convertible = std::is_convertible<T,U>::value && std::is_convertible<U,T>::value;
~~~~~~~~~~~~~~~~~~~~~~~~~~^
I've tried
g++-5.4, g++-6.2 (no error)
clang++-3.5, clang++-3.8, clang++-3.9 (error)
with argument -std=c++1y and for clang either with -stdlib=libstdc++ or -stdlib=libc++.
Which compiler is correct? Is it a bug in clang or gcc? Or is the behavior for some reasons undefined and thus both compilers correct?

Answer 1

First of all, note that it works fine if you use:
A<double> s{1.0};
Instead, the error comes from the fact that you are doing this:
A<double> s = 1.0;
Consider the line below (extracted from the definition of Convertible):
std::is_convertible<U,T>::value
In your case, this is seen as it follows (once substitution has been performed):
std::is_convertible<double, A<double>>::value
The compiler says this clearly in the error message.
This is because a temporary A<double> is constructed from 1.0, then it is assigned to s.
Note that in your class template you have defined a (more or less) catch-all constructor, so a const A<double> & is accepted as well.
Moreover, remember that a temporary binds to a const reference.
That said, the error happens because in the context of the std::enable_if_t we have that A<double> is an incomplete type and from the standard we have this for std::is_convertible:
From and To shall be complete types [...]
See here for the working draft.
Because of that, I would say that it's an undefined behavior.
As a suggestion, you don't need to use std::enable_if_t in this case.
You don't have a set of functions from which to pick the best one up in your example.
A static_assert is just fine and error messages are nicer:
template <class S>
A(S const&) { static_assert(Convertible<S,T>, "!"); }