模板类中的静态成员

Question

I have a templated class and a static member of it, which is of type map. I can't seem to figure out how to use that member. I have tried a few different variations but this is where it began.
I get an error in the definition of function size() :

undefined reference to `A<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >::my_map_'    




#include <iostream>
#include <map>
#include <string>

using namespace std;

typedef map<string, int> si_map_t;

template<class T>
class A {
public:
    static map<T, int> my_map_;
    static void add(T key, int value) {
        my_map_.insert(std::pair<T, int>(key, value));
    }
    static size_t size() {
        return my_map_.size();
    }
};

template<>
si_map_t A<string>::my_map_;



int main() {
    A<string>::my_map_;
    size_t count = A<string>::size();
    cout << count << endl;
    return 0;
}

Answer 1

I think this is what you're trying to do:

template<typename T>
using si_map_t = std::map<T, int>;

template<class T>
class A {
public:
    static si_map_t<T> my_map_;
    static void add(T key, int value) {
        my_map_.insert(std::pair<T, int>(key, value));
    }
    static size_t size() {
        return my_map_.size();
    }
};

template<typename T>
si_map_t<T> A<T>::my_map_;

int main()
{
    size_t count = A<string>::size();
    cout << count << endl;
    return 0;
}

Output

0

Answer 2

If you really want to use explicit specialization of a class template member note that such notation

template <>
si_map_t A<string>::my_map_;

means only a declaration of a specialization. It is strange but it is true.

Use the following syntax to define a member specialization:

template <>
si_map_t A<string>::my_map_ = si_map_t();

Or the following C++11 syntax:

template <>
si_map_t A<string>::my_map_ = {};

or

template <>
si_map_t A<string>::my_map_{};