Does C ++ 11 support template class reflection?

Question

I know a little bit of knowledge about the C ++ 11 template. My intention is to have a template function as shown below:

template<class T>
void function(T * a) {
  if (T belongs to class M) {
    a->function_m();
  } else {
    a->function_o();
  }
}

Does C ++ 11 support this reflection of the template class?

Answer 1

Yes, and even better, you do not need to follow the instructions if(...){} else{}for this. You can use tag manager or specialization to avoid conditional statements. The following example uses tag dispatch.

Example:

#include <iostream>
#include <type_traits>

template <typename B, typename D>
void function( D* a )
{
    function( a, typename std::is_base_of<B, D>::type{} );
}

template <typename T>
void function( T* a, std::true_type )
{
    a->function_b();
}

template <typename T>
void function( T* a, std::false_type )
{
    a->function_c();
}

struct B
{
    virtual void function_b() { std::cout << "base class.\n"; }
};

struct D : public B
{
    void function_b() override { std::cout << "derived class.\n"; }
};

struct C
{
    void function_c() { std::cout << "some other class.\n"; }
};

int main()
{
    D d;
    C c;
    function<B, D>( &d );
    function<B, C>( &c );
}

This mechanism does not require both functions to be visible in the same area.

Answer 2

Several variants:

• SFINAE:

  template<class T>
  std::enable_if_t<std::is_base_of<M, T>>
  function(T* a)
  {
      a->function_m();
  }
  
  template<class T>
  std::enable_if_t<!std::is_base_of<M, T>>
  function(T* a)
  {
      a->function_o();
  }
  

• or sending tags:

  namespace details {
      template<class T>
      void function(T* a, std::true_type) {
          a->function_m();
      }
  
      template<class T>
      void function(T* a, std::false_type) {
          a->function_o();
      }
  }
  template<class T>
  void function(T* a)
  {
      details::function(a, std::is_base_of<M, T>{});
  }
  

Answer 3

, std::is_base_of<Base,Derived>:

template<class T>
void function(T * a) {
  if (std::is_base_of<M,T>::value) {
    a->function_m();
  } else {
    a->function_o();
  }
}

However, probably, it will cause a problem in this case, as function_m()and function_o()should have been called.

Answer 4

What do you want to do in C ++ 17

template <typename T>
void function( T* a )
{
    if constexpr (std::is_base_of<M,T>::value)
        a->function_m();
    else
        a->function_o();
}

Full example: http://melpon.org/wandbox/permlink/MsHnYQNlBcRhTu2C As mentioned by @Fabio Fracassi