Question

How can I specialize or overload function func so that specialization handles all instances of MyClass? Assume func is a librarian function (like std::swap for example) so I can’t change func, and I can’t replace or wrap it, I have to specialize it.

#include <iostream>

template<typename T>
class MyClass
{
};

template<typename T>
void func(const T&)
{
    std::cout << "Default" << std::endl;
}

// I don't want to copy this stuff for every instance of MyClass
template<>
void func<MyClass<int>>(const MyClass<int>&)
{
    std::cout << "Specialization" << std::endl;
}

int main(int, char**)
{
    func(int(0)); // "Default"

    func(MyClass<int>()); // "Specialization"

    func(MyClass<double>()); // "Default" but I want "Specialization" here

    return 0;
}

Answer 1

Here are a few examples of what you can do.

template<typename T>
void func(const T&)
{
    std::cout << "Default" << std::endl;
}

template<template<typename> class A , typename B>
void func(const A<B>&)
{
    std::cout << "Overload All templated classes with one templated argument B" << std::endl;
}

template<typename T>
void func(const MyClass<T>&)
{
    std::cout << "Overload Myclass with some templated argument T" << std::endl;
}