Home/ Questions/Q 6095627
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-23T12:49:59+00:00

While reading this question , I came across @Johannes ‘s answer. template<typename> struct void_

  • 0

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 ?
Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    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.

    • 0