SFINAE not working with member function of template class

Question

I have a template class where I would like to remove a member function if the type satisfies some condition, that, as far as I understand, should be a very basic usage of SFINAE, for example:

template<class T>
class A
{
public:
    template<typename = typename std::enable_if<std::is_floating_point<T>::value>::type>
    T foo () {
        return 1.23;
    }
};

However, this is results in an error "no type named 'type'", like SFINAE was not going on. This however works if foo is a function not member of a class. What is wrong with this implementation?

Answer 1

You're missing a dependent name for the compiler to use for SFINAE. Try something like this instead:

#include <type_traits>

template<class T>
class A
{
public:
    template<typename Tp = T>
    typename std::enable_if<std::is_floating_point<Tp>::value, Tp>::type
    foo () {
        return 1.23;
    }
};

int main() {
    A<double> a;
    a.foo();
}

If the type T is not floating point, the declaration would be malformed (no return type) and the function would not be considered for the overload set.

See it on godbolt.