C++ Template overload with enable_if: different behaviour with g++ and clang - c++

Question

During resolution of an overload of a templated member function of a base class, I observed a different behaviour between g++ (5.2.1-23) and clang (3.8.0), with -std=c++14.
#include <iostream>
#include <type_traits>
struct Base
{
template <typename T>
auto a(T t) -> void {
std::cout<< "False\n";
}
};
template <bool Bool>
struct Derived : public Base
{
using Base::a;
template <typename T, bool B = Bool>
auto a(T t) -> std::enable_if_t<B, void>
{
std::cout<< "True\n";
}
};
int main()
{
Derived<true> d;
d.a(1); // fails with g++, prints "true" with clang
Derived<false> d2;
d2.a(1); // fails with clang++, prints "false" with g++
}
The call to Derived<true>::a fails with g++ with the following message:
test.cc: In function ‘int main()’:
test.cc:28:8: error: call of overloaded ‘a(int)’ is ambiguous
d.a(1);
^
test.cc:18:8: note: candidate: std::enable_if_t<B, void> Derived<Bool>::a(T) [with T = int; bool B = true; bool Bool = true; std::enable_if_t<B, void> = void]
auto a(T t) -> std::enable_if_t<B, void>
^
test.cc:7:8: note: candidate: void Base::a(T) [with T = int]
auto a(T t) -> void {
^
and the call to Derived<false>::a fails with clang++ with the following message:
test.cc:32:6: error: no matching member function for call to 'a'
d2.a(1);
~~~^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.2.1/../../../../include/c++/5.2.1/type_traits:2388:44: note: candidate template ignored: disabled by 'enable_if' [with T = int, B = false]
using enable_if_t = typename enable_if<_Cond, _Tp>::type;
^
My guess is that they interpret differently the using Base::a;, and that it isn't considered in clang, whereas it's (maybe too much) considered in g++. What I thought would happen is that if Derived has true as parameter, then the call of a() is dispatched to Derived's implementation, whereas if the parameter is false, the call is dispatched to Base::a.
Are they both wrong? Who is right? Who should I submit a bug report to? Can somebody explain what is going on?
Thanks

Answer 1

From 3.3.10/p3 Name hiding [basic.scope.hiding]:
In a member function definition, the declaration of a name at block
scope hides the declaration of a member of the class with the same
name; see 3.3.7. The declaration of a member in a derived class
(Clause 10) hides the declaration of a member of a base class of the
same name; see 10.2
Also 7.3.3/p15 The using declaration [namespace.udecl]:
When a using-declaration brings names from a base class into a derived
class scope, member functions and member function templates in the
derived class override and/or hide member functions and member
function templates with the same name, parameter-type-list (8.3.5),
cv-qualification, and ref-qualifier (if any) in a base class (rather
than conflicting). [ Note: For using-declarations that name a
constructor, see 12.9. — end note ] [Example:
struct B {
virtual void f(int);
virtual void f(char);
void g(int);
void h(int);
};
struct D : B {
using B::f;
void f(int); // OK: D::f(int) overrides B::f(int);
using B::g;
void g(char); // OK
using B::h;
void h(int); // OK: D::h(int) hides B::h(int)
};
void k(D* p)
{
p->f(1); // calls D::f(int)
p->f(’a’); // calls B::f(char)
p->g(1); // calls B::g(int)
p->g(’a’); // calls D::g(char)
}
— end example ]
This is resolved during member name look-up. Thus, it's before template argument deduction. Consequently, as correctly TC mentioned in the comments Base template function is hidden no matter of SFINAE verdict.
Therefore CLANG is correct and GCC is wrong.