Question

With a template method, that is specialized for a specific class A, is there a way to call the specialized code for the classes deriving from A ?

The minimal example

#include <iostream<  

class A
{};

class B : public A
{};

class Templator
{
public :
    template <class T>
    void dance(T *argument);
};

template <class T>
void Templator::dance(T *argument)
{
    std::cout << "General implementation" << std::endl;
}

template<>
void Templator::dance<A> (A* )
{
    std::cout << "Specialized implementation" << std::endl;
}

int main()
{
    A a;
    B b;

    Templator tt;
    tt.dance(&a);
    tt.dance(&b);

    return 0;
}

Corresponding output

Specialized implementation
General implementation

In other words, I would like the two outputs to be :

Specialized implementation
Specialized implementation

Answer 1

In C++11 you can use std::enable_if and std::is_base_and_derived:

class Templator{
public :
    template <class T>
    typename std::enable_if<!std::is_base_and_derived<A, T>::value, void>::type 
    dance(T *argument) { 
       std::cout << "General implementation" << std::endl;
    }
    template <class T>
    typename std::enable_if<std::is_base_and_derived<A, T>::value, void>::type 
    dance(T *argument) { 
       std::cout << "Specialized implementation" << std::endl;
    }
};

You can do the same thing in C++03 using either boost::enable_if_c and boost::is_base_of or std::tr1::is_base_of and writing enable_if yourself (see here)