c++ - SFINAE 不能防止模棱两可的运算符重载吗？

Question

我认为下面的代码会编译，因为冲突的重载是 SFINAEd 了。但是编译器(GCC)说:void Foo<X>::bar(Xd) const' cannot be overloaded .有没有简单的方法来修复它，或者我必须专门研究 Foo ?

#include <type_traits>

struct A{};

template<typename X>
struct Foo {
    template<typename Xd=X, typename = std::enable_if_t<std::is_arithmetic<Xd>::value>>
    void bar() const {
    }

    template<typename Xd=X, typename = std::enable_if_t<std::is_same<Xd,A>::value>>
    void bar() const {
    }
};

int main(){}

Answer 1

来自 the reference对于 std::enable_if:

A common mistake is to declare two function templates that differ only in their default template arguments. This does not work because the declarations are treated as redeclarations of the same function template (default template arguments are not accounted for in function template equivalence).

/*** WRONG ***/
struct T {
    enum { int_t,float_t } m_type;
    template <typename Integer,
              typename = std::enable_if_t<std::is_integral<Integer>::value>
    >
    T(Integer) : m_type(int_t) {}

    template <typename Floating,
              typename = std::enable_if_t<std::is_floating_point<Floating>::value>
    >
    T(Floating) : m_type(float_t) {} // error: treated as redefinition
};


/* RIGHT */   
struct T {
    enum { int_t,float_t } m_type;
    template <typename Integer,
              std::enable_if_t<std::is_integral<Integer>::value, int> = 0
    >
    T(Integer) : m_type(int_t) {}

    template <typename Floating,
              std::enable_if_t<std::is_floating_point<Floating>::value, int> = 0
    >
    T(Floating) : m_type(float_t) {} // OK
};