使用std :: is_same进行元编程

Question

Is it possible to do something like the following that compiles without template specialization?

template <class T> 
class A {
public:
  #if std::is_same<T,int>
  void has_int() {}
  #elif std::is_same<T,char>
  void has_char() {}
  #endif
};
A<int> a; a.has_int();
A<char> b; b.has_char();

Answer 1

Yes. Make the function templates and then conditionaly enable them using std::enable_if :

#include <type_traits>

template <class T> 
class A {
public:

  template<typename U = T>
  typename std::enable_if<std::is_same<U,int>::value>::type
  has_int() {}

  template<typename U = T>
  typename std::enable_if<std::is_same<U,char>::value>::type
  has_char() {}
};

int main()
{
    A<int> a;
    a.has_int();   // OK
    // a.has_char();  // error
}

The solution from the other answer might not be feasible if the class is big and has got many functions that need to regardless of T . But you can solve this by inheriting from another class that is used only for these special methods. Then, you can specialize that base class only.

In C++14, there are convenient type aliases so the syntax can become:

std::enable_if_t<std::is_same<U, int>::value>

And C++17, even shorter:

std::enable_if_t<std::is_same_v<U, int>>

Answer 2

Yes, with template specialization :

template <class T> 
class A;

template <> 
class A<int>
{
    void had_int(){}
};

template <> 
class A<char>
{
    void had_char(){}
};