Is an is_functor C++ trait class possible?

Question

How can I deduce statically if an argument is a C++ function object (functor)?

template 
void test(F f) {}

I tried

Answer 1

Although this does not work for overloaded functions, for all other cases (free functions, classes implementing operator(), and lambdas) this short solutions works in C++11:

template 
struct is_callable: std::is_convertible> { };

Note: std::is_invocable is available since C++17.