c++ - Different template syntax for finding if argument is a class or not

Question

While reading this question , I came across @Johannes's answer.

template<typename> struct void_ { typedef void type; };

template<typename T, typename = void>  // Line 1
struct is_class { static bool const value = false; };

template<typename T>
struct is_class<T, typename void_<int T::*>::type> { // Line 2
  static bool const value = true; 
};

This construct finds if the given type is a class or not. What puzzles me is the new kind of syntax for writing this small meta program. Can anyone explain in detail:

1. Why we need Line 1 ?
2. What is the meaning of syntax <int T::*> as template parameter in Line 2 ?

Answer 1

Line 1: Choosing the partial specialization below if the test succeeds.

Line 2: int T::* is only valid if T is a class type, as it denotes a member pointer.

As such, if it is valid, void_<T>::type yields void, having this partial specialization chosen for the instantiation with a value of true. If T is not of class type, then this partial specialization is out of the picture thanks to SFINAE and it defaults back to the general template with a value of false.

Everytime you see a T::SOMETHING, if SOMETHING isn't present, be it a type, a data member or a simple pointer definition, you got SFINAE going.