C ++：用于operator <的模板类中的friend函数

Question

What is the correct way to declare a friend function of a template class (for the std::ostream& operator<<) in a .cpp file ?

My current implementation does not work :

// MyTest.h
template<class T, unsigned int TSIZE> class MyTest
{
    inline friend std::ostream& operator<< <T, TSIZE> (std::ostream &lhs, const MyTest<T, TSIZE> &rhs);
};

// MyTest.cpp
template<class T, unsigned int TSIZE> inline friend std::ostream& operator<< <T, TSIZE> (std::ostream &lhs, const MyTest<T, TSIZE> &rhs)
{
    // IMPLEMENTATION
}

Thank you very much !

Answer 1

To refer to operator<< <T, TSIZE> like you do, which is a template specialization, a declaration of the primary template must be visible. In turn operator<< needs a declaration of MyTest because it appears as a parameter.

// Declare MyTest because operator<< needs it
template<class T, unsigned int TSIZE> class MyTest;

// Declare primary template
template<class T, unsigned int TSIZE>
inline std::ostream& operator<<(std::ostream& lhs, const MyText<T, TSIZE>& rhs);

template<class T, unsigned int TSIZE> class MyTest
{
    // Specialization MyTest<T, TSIZE> only declares
    // specialization operator<< <T, TSIZE> as friend
    // Note that you can just use '<>' to designate the specialization,
    // template parameters are deduced from the argument list in this case
    inline friend std::ostream& operator<< <> (std::ostream &lhs, const MyTest<T, TSIZE> &rhs);
};

The definition you have should match those declarations. Note that since operator<< is a template its definition should in all likeliness be in the header.

An alternative that requires less work when it comes to writing all those preemptive declarations is for MyTest<T, TSIZE> to declare the whole template as a friend, not just the specialization that takes MyTest<T, TSIZE> .

// in MyTest definition
template<typename U, unsigned USIZE>
inline friend std::ostream& operator<<(std::ostream& lhs, const MyTest<U, USIZE>& rhs);

The definition you have should also match such a declaration (the name of the template parameters has no bearing on matching declarations and definition).

For the sake of completeness, I will mention that when it comes to the friend of a class template an alternative is to define it in the class template definition. This defines a non-template friend function that is unique for each specialization.

// in MyTest definition
friend std::ostream& operator<<(std::ostream& lhs, MyTest const& rhs)
{ /* implementation */ }

It is impossible to refer to such functions (eg &ns::operator<< doesn't work, unlike the other options) and they are only found via ADL.

Answer 2

It is not fully clear what the original post wanted. I will assume that it wanted the following:

// Some template class.
template<class T, unsigned int TSIZE> class MyTest { };

// Some template function.
template<class T, unsigned int TSIZE> std::ostream& operator<< (std::ostream &lhs, const MyTest<T, TSIZE> &rhs)
{
    // IMPLEMENTATION
}

Now it is necessary to declare this template function as friend of the class because this function needs access to protected objects of My test . This can be achieved with the following definition:

template<class T, unsigned int TSIZE> class MyTest
{
    template<class T1, unsigned int TSIZE1> 
    friend std::ostream& operator<< (std::ostream &lhs, const MyTest<T1, TSIZE1> &rhs);
};

Template header is needed in front of the friend declaration because this is an unrelated template function, that does not belong to the current template class.