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 t

Answer 1

No, but you can overload with a template function which only accepts std::vector<T>. The compiler will choose the most specialized template in such cases.

Answer 2

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;
}

Answer 3

If you need a trait class it's pretty simple, you only need a general template and a specialization over any std::vector:

#include <type_traits>
#include <iostream>

template<typename>
struct is_std_vector : std::false_type {};

template<typename T, typename A>
struct is_std_vector<std::vector<T,A>> : std::true_type {};

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