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; } 

Answer 2

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> #include <vector>  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; }