I am trying to declare a function as friend to a class template with a protected member. Below is a minimal example.
template<int N> class myClass{
public:
friend void f(const myClass& c);
protected:
int a;
};
If I now define the function as
template<int N> void f(const myClass<N>& c){
std::cout << c.a;
};
Then it works.
However if I use template specialisation
template<int N> void f(const myClass<N>& c);
template<> void f<1>(const myClass<1>& c){
std::cout << c.a;
};
It does not recognise f as a friend anymore and complains that a is a protected member.
Why is that happening? What am I doing wrong?
CodePudding user response：
The problem is that the friend declaration friend void f(const myClass& c); is a nontemplate friend declaration. That is, you're actually befriending a nontemplae free function. This is exactly what the warning tells you:
warning: friend declaration 'void f(const myClass<N>&)' declares a non-template function [-Wnon-template-friend]
12 | friend void f(const myClass& c);
To solve this you need to add a separate parameter clause for the friend declaration as shown below:
template<int N> class myClass{
public:
template<int M> //added this parameter clause
friend void f(const myClass<M>& c);
protected:
int a;
};