Ambiguous operator<< selection - c++11

Question

I have some code which, very much simplified, looks somewhat like this:
#include <iostream>
#include <type_traits>
namespace X {
struct Foo {int x;};
struct Bar {int x;};
template <typename T , typename = typename std::enable_if<
std::is_same<decltype(T::x),int>::value
>::type>
std::ostream & operator<<(std::ostream & os, const T&) {
return os;
}
}
namespace Y {
struct Faa : X::Foo {int y;};
struct Baz {int x; int y;};
template <typename T , typename = typename std::enable_if<
std::is_same<decltype(T::x),int>::value &&
std::is_same<decltype(T::y),int>::value
>::type>
std::ostream & operator<<(std::ostream & os, const T&) {
return os;
}
}
int main() {
// Everything is ok
X::Foo x;
std::cout << x;
Y::Baz k;
std::cout << k;
// Problems..
Y::Faa y;
// std::cout << y; // <--operator is ambiguous
Y::operator<<(std::cout, y);
return 0;
}
Is there any way to avoid the ambiguous operator for Y::Faa and having to manually specify Y::operator<<? If not, why?

Answer 1

Two functions have a conflict because conditions on their arguments have non-empty intersection (actually, 1st supersedes 2nd). Function overloading works only if signatures are different. So, to solve this we have 2 options:
Change conditions so that they have empty intersection (manually forbid having y field by adding && !sfinae_has_member_y<T>::value condition to the 1st enable_if)
template<typename T>
struct sfinae_has_member_y {
static int has(...);
template<typename U = T, typename = decltype(U::y)>
static char has(const U& value);
enum { value = sizeof(char) == sizeof(has(std::declval<T>())) };
};
OR use another C++ feature that supports arguments overlapping, like struct/class template specialization. If you replace bool with int, other fields may be added too:
template<typename T, bool>
struct Outputter {
};
template<typename T>
struct Outputter<T, false> {
static std::ostream & output(std::ostream & os, const T&) {
os << "x";
return os;
}
};
template<typename T>
struct Outputter<T, true> {
static std::ostream & output(std::ostream & os, const T&) {
os << "y";
return os;
}
};
template<typename T, typename = std::enable_if_t<std::is_same<decltype(T::x), int>::value>>
std::ostream & operator<<(std::ostream & os, const T& a) {
return Outputter<T, sfinae_has_member_y<T>::value>::output(os, a);
}