Question

I need to create a template function like this:

template<typename T>
void foo(T a)
{
   if (T is a subclass of class Bar)
      do this
   else
      do something else
}

I can also imagine doing it using template specialization … but I have never seen a template specialization for all subclasses of a superclass. I don’t want to repeat specialization code for each subclass

Answer 1

You can do what you want but not how you are trying to do it! You can use std::enable_if together with std::is_base_of:

#include <iostream>
#include <utility>
#include <type_traits>

struct Bar { virtual ~Bar() {} };
struct Foo: Bar {};
struct Faz {};

template <typename T>
typename std::enable_if<std::is_base_of<Bar, T>::value>::type
foo(char const* type, T) {
    std::cout << type << " is derived from Bar\n";
}
template <typename T>
typename std::enable_if<!std::is_base_of<Bar, T>::value>::type
foo(char const* type, T) {
    std::cout << type << " is NOT derived from Bar\n";
}

int main()
{
    foo("Foo", Foo());
    foo("Faz", Faz());
}

Since this stuff gets more wide-spread, people have discussed having some sort of static if but so far it hasn’t come into existance.

Both std::enable_if and std::is_base_of (declared in <type_traits>) are new in C++2011. If you need to compile with a C++2003 compiler you can either use their implementation from Boost (you need to change the namespace to boost and include "boost/utility.hpp" and "boost/enable_if.hpp" instead of the respective standard headers). Alternatively, if you can’t use Boost, both of these class template can be implemented quite easily.