Is this considered SFINAE?

Question

I asked a question about a week ago inquiring how I would be able to simply instantiate a class template only if the type it took had a specific member function. In my answer I got sort of a complicated solution. But then I tried to do it on my own. I just wanted to know if this enough to figure out of a given type T has a void function named f taking 0 parameters.

#include <type_traits>
#include <utility>

template <typename T, typename = void>
struct has_f : std::false_type { };

template <typename T>
struct has_f<
    T,
    decltype(std::declval<T>().f(), void())> : std::true_type { };

template <typename T, typename = typename std::enable_if<has_f<T>::value>::type>
struct A { };

struct B
{
    void f();
};

struct C { };

template class A<B>; // compiles
template class A<C>; // error: no type named ‘type’ 
                     // in ‘struct std::enable_if<false, void>’

If so, why are the other answers so complicated in this thread?

Answer 1

Yep, you have solved it in the simplest, most idiomatic style of C++11 SFINAE.

Note that you didn't check that the return type is void, that it's a nonstatic member, nor that there are no parameters. f is simply callable with no arguments. It could even be a functor.

To check for a nullary member nonstatic function returning void, use

template <typename T>
struct has_f<T, decltype(void( static_cast< void (T::*)( void ) >( &T::f ) )) >
    : std::true_type {};

template <typename T>
struct has_f<T, decltype(void( static_cast< void (T::*)( void ) const >( &T::f ) )) >
    : std::true_type {};