how to overload operator == outside template class

Question

I'm trying to write a template class which overloads operator==. I know how to get it inside the class:

    template <typename T>
    class Point
    {
    private:
        T x;
    public:
        Point(T X) : x(X) {}

        bool operator== (Point &cP)
        {
            return (cP.x == x);
        }
    };

But now I want to achieve this outside the template class. I've read this post: error when trying to overload << operator and using friend function and add template declaration in my code:

template <typename> class Point;
template <typename T> bool operator== (Point<T>, Point<T>);
template <class T>
class Point
{
private:
    T x;
public:
    Point(T X) : x(X) {}

    friend bool operator== (Point cP1, Point cP2);
};

template <class T>
bool operator== (Point<T> cP1, Point<T> cP2)
{
    return (cP1.x == cP2.x)
}

However I still get a error: unresolved external symbol "bool __cdecl operator==(class Point<int>,class Point<int>)" (??8@YA_NV?$Point@H@@0@Z) referenced in function _main

And when I take away friend from :

friend bool operator== (Point cP1, Point cP2);

and want it to be member function, there would be a another error:

too many parameters for this function

why?

Answer 1

You have to put friend classes within each other, as well as in themselves. The Real one is named syntactically the same without replacing expressions. So friend bool operator==(P<T>,P<T>); goes in both of the classes. So put friend before bool in the class T :-)

Answer 2

This is a tricky issue: the friend declaration in the class will define a different friend function for each instantiation of your class. In other words, Point<int> and Point<double> will give rise to 2 different non-template friend functions. On the other hand, the outside function is a template function, not related to the friend inside the class. Solution: define the friend function inside the class. Due to friend name injection, it will be found successfully by ADL.

Answer 3

@Kühl's answer is the most permissive approach to declare a templated friend function of a templated class. However, there is one unapparent side effect of this approach: All template instantiations of Point are friends with all template instantiations of operator==(). An alternative is to make only the instantiation with the same type of Point a friend. Which is done by adding a <T> in the friend declaration of operator==().

template <typename T> class Point;

template <typename S>
bool operator== (Point<S>, Point<S>);

template <typename T>
class Point {
    // ...
    friend bool operator==<T> (Point, Point);
};

References
http://web.mst.edu/~nmjxv3/articles/templates.html

Answer 4

The declaration of the operator==() is a template. The declaration made a friend is not a template but a non-template. To make the template operator==() a friend you need to make the friend declaration a template, too:

template <typename T> class Point;

template <typename S>
bool operator== (Point<S>, Point<S>);

template <typename T>
class Point {
    // ...
    template <typename S>
    friend bool operator== (Point<S>, Point<S>);
};