template friend functions of template class

Question

I have the following template class and template function which intends to access the class\' private data member:

#include 

template

        

Answer 1

The simplest option is to define the friend within the class:

template<class T>
class MyVar
{
    int x;

    friend void printVar(const MyVar & var) {
        std::cout << var.x << std::endl;
    }
    friend void scanVar(MyVar & var) {
        std::cin >> var.x;
    }
};

The downside is that the functions can only be called through argument-dependent lookup. That's not a problem in your example, but might be a problem if they don't have a suitable argument, or you want to specify the name without calling it.

If you want a separate definition, then the template will have to be declared before the class definition (so it's available for a friend declaration), but defined afterwards (so it can access the class members). The class will also have to be declared before the function. This is a bit messy, so I'll only show one of the two functions:

template <typename T> class MyVar;
template <typename T> void printVar(const MyVar<T> & var);

template<class T>
class MyVar
{
    int x;

    friend void printVar<T>(const MyVar<T> & var);
};

template <typename T> void printVar(const MyVar<T> & var) {
    std::cout << var.x << std::endl;
}

Answer 2

I managed to get the following work

#include <iostream>

template<class T>
class MyVar;

template<class T>
void printVar(const MyVar<T>& var);

template<class T>
void scanVar(MyVar<T>& var);

template<class T>
class MyVar
{
    int x;
    friend void printVar<T>(const MyVar<T>& var);
    friend void scanVar<T>(MyVar<T>& var);
};

template<class T>
void printVar(const MyVar<T>& var)
{
    std::cout << var.x << std::endl;
}

template<class T>
void scanVar(MyVar<T>& var)
{
    std::cin >> var.x;
}

struct Foo {};

int main(void)
{
    MyVar<Foo> a;
    scanVar(a);
    printVar(a);
    return 0;
}

UPD: http://en.cppreference.com/w/cpp/language/friend talks about a similar case with operators under "Template friend operators":

A common use case for template friends is declaration of a non-member operator overload that acts on a class template, e.g. operator<<(std::ostream&, const Foo<T>&) for some user-defined Foo<T>

Such operator can be defined in the class body, which has the effect of generating a separate non-template operator<< for each T and makes that non-template operator<< a friend of its Foo<T>

...

or the function template has to be declared as a template before the class body, in which case the friend declaration within Foo<T> can refer to the full specialization of operator<< for its T

Answer 3

This one compiles on MSVC2013. Basicly adds the forward declarations to class and functions before the friend

template<class T>   class MyVar ; // class forward declaration

template<class T> ; // function forward declarations
void printVar(const MyVar<T>& var);
template<class T>
void scanVar(MyVar<T>& var);

template<class T>
class MyVar
{
    friend void printVar<T>(const MyVar<T>&);
    friend void scanVar<T>(MyVar<T>&);
    int x;
};

template<class T>
void printVar(const MyVar<T>& var)
{
    std::cout << var.x << std::endl;
}

template<class T>
void scanVar(MyVar<T>& var)
{
    std::cin >> var.x;
}

struct Foo {};

int main1(void)
{
    MyVar<Foo> a;
    scanVar(a);
    printVar(a);
    return 0;
}

Answer 4

Well there is a solution that is both simple and involving separation between declaration & definition of the friend function. In the declaration of the friend function (inside the class) you have to give a different template param from the one the class accepts (and it make sense cause this function is not a member of this class).

template<class T>
class MyVar
{
    int x;

    template<typename Type>
    friend void printVar(const MyVar<Type> & var);

    template<typename Type>
    friend void scanVar(MyVar<Type> & var);
};

template<typename T>
void printVar(const MyVar<T> & var) {

}

template<typename T>
void scanVar(MyVar<T> & var) {

}

No forward declaration needed, as well as there is a separation of declaration & definition.