How to specialize for template template arguments

Question

I would like to specialize a function using a templated type but I am having trouble getting the desired results.

Consider the following simple example

#include <iostream>

#include <typeinfo>

#include <vector>





template <typename T>

void foo(){

    std::cout << "In foo1 with type: " << typeid(T).name() << std::endl;

}



template< template<class, class...> class VEC, typename T>

void foo(){

    std::cout << "In foo2 with vec type: " << typeid(VEC<T>).name()

              << " and template type: " << typeid(T).name() << std::endl;

}





int main() {

    foo<int>();

    foo<std::vector, int>();

    foo<std::vector<int>>(); // Would like this to call the second version of foo

}

The output of which is

In foo1 with type: i

In foo2 with vec type: St6vectorIiSaIiEE and template type: i

In foo1 with type: St6vectorIiSaIiEE

Is there a way to write a template signature for the second version of foo that would be used for the last invocation of foo (with the std::vector template parameter)?

Thanks!

Recommended for you: Get network issues from WhatsUp Gold. Not end users.

Answer 1

Since you cannot specialize function templates partially, the usual approach is to use a helper class template:

template <typename T> struct X
{
    static void f() { std::cout << "Primary\n"; }
};

template <template <typename...> class Tmpl, typename T>
struct X<Tmpl<T>>
{
    static void f() { std::cout << "Specialized\n"; }
};

template <typename T> void foo() { X<T>::f(); }

How to specialize for template template arguments

Sign up