How to specialize template function with template types

Question

Is it possible to specialize a template function for template types? I don't know if my terminology is correct, so I'll provide a simple sample with what I want to achieve:

#include <vector>
#include <string>
#include <iostream>

template<typename T>
void f()
{
    std::cout << "generic" << std::endl;
}

template<>
void f<std::string>()
{
    std::cout << "string" << std::endl;
}

template<typename T>
void f<std::vector<T>>()
{
    std::cout << "vector" << std::endl;
}

int main()
{
    f<double>();
    f<std::string>();
    f<std::vector<int>>();

    return 0;
}

This code doesn't compile. VS2013 gives me

error C2995: 'void f(void)' : function template has already been defined

on this function:

template<typename T>
void f<std::vector<T>>()
{
    std::cout << "vector" << std::endl;
}

How may I achieve this behavior? It's very important to have type f(void) signature. Is this code falling into partial specialization for functions(forbidden in C++)?

Answer 1

You can't partially specialize template function, but you can for template class. So you can forward your implementation to a dedicated class. Following may help: (https://ideone.com/2V39Ik)

namespace details
{
    template <typename T>
    struct f_caller
    {
        static void f() { std::cout << "generic" << std::endl; }
    };

    template<>
    struct f_caller<std::string>
    {
        static void f() { std::cout << "string" << std::endl; }
    };

    template<typename T>
    struct f_caller<std::vector<T>>
    {
        static void f() { std::cout << "vector" << std::endl; }
    };
}

template<typename T>
void f()
{
    details::f_caller<T>::f();
}