具有 cv 和 ref 类型的模板特化

Question

In order to understand metaprogramming I created a simple example:

template <class T> struct add_cref      { typedef T const& type; };
// template <class T> struct add_cref<T&>   { typedef T const& type; };


std::cout << std::boolalpha
    << std::is_same<add_cref<int>::type, int const&>::value << std::endl
    << std::is_same<add_cref<int&>::type, int const&>::value << std::endl
    << std::is_same<add_cref<const int>::type, int const&>::value << std::endl
    << std::is_same<add_cref<int const&>::type, int const&>::value << std::endl;

The result is: true, false, true, true
When I uncomment the template spec, the result is as expected (all true)

My question is, why is the second one false and the last one true without the specialization when both use the specialization when uncommented.

Answer 1

template <class T> 
struct add_cref { 
    typedef T const& type; 
};

With the type add_cref<int&>::type , T = int& . The type add_cref<int&>::type is then roughly the same as int& const & , which means that the reference int& is const and not the integer itself.

Edit: With the type add_cref<const int&>::type , T = const int& . The type add_cref<const int&>::type is then roughly the same as const int& const & , which means that the reference itself const int& is const (the second const is ignored by the compiler) but it refers to a const int . This means that add_cref<const int&>::type must be const int& , even without specialization.

With specialization:

template <class T> 
struct add_cref<T&> { 
   typedef T const& type; 
}; 

For add_cref<int&> since in this specialization T&=int& then T=int . As a result, the type of T const& becomes int const & .