How to know if a type is a specialization of std::vector?

Question

I've been on this problem all morning with no result whatsoever. Basically, I need a simple metaprogramming thing that allows me to branch to different specializations if the parameter passed is a kind of std::vector or not.

Some kind of is_base_of for templates.

Does such a thing exist ?

Answer 1

In C++11 you can also do it in a more generic way:

#include <type_traits>
#include <iostream>
#include <vector>
#include <list>

template<typename Test, template<typename...> class Ref>
struct is_specialization : std::false_type {};

template<template<typename...> class Ref, typename... Args>
struct is_specialization<Ref<Args...>, Ref>: std::true_type {};


int main()
{
    typedef std::vector<int> vec;
    typedef int not_vec;
    std::cout << is_specialization<vec, std::vector>::value << is_specialization<not_vec, std::vector>::value;

    typedef std::list<int> lst;
    typedef int not_lst;
    std::cout << is_specialization<lst, std::list>::value << is_specialization<not_lst, std::list>::value;
}