Question

Say I have two Template classes.

template<class T>
class baseclass1
{
    template<class> friend class baseclass2;
}

template<class D>
class baseclass2
{
     template<class T> void foo( D& x, T& y)
     {
          ...
     }
}

The Above code allows all types of baseclass1 to friend all types of baseclass2, a many-to-many relationship. I have two questions,

What is the syntax to allow baseclass1 to friend just the function

baseclass2<class D>::foo<class T>( D& x, T& y).  

And, what is the syntax to allow baseclass1 to friend just the function

baseclass2<class D>::foo<class T>( D& x, T& y) where T from baseclass1 matches The T from Function foo.

EDIT

To those who keep claiming you can’t friend a template specialization. This code works

template<class cake>
class foo
{
    public:
        static void bar(cake x)
        {
            cout << x.x;
        }
};


class pie
{
    public:
        void set( int y){ x = y; }
    private:
        int x;

        friend void foo<pie>::bar(pie x);
};

class muffin
{
    public:
        void set( int y){ x = y; }
    private:
        int x;

    friend void foo<pie>::bar(pie x);
};

int main
{
        pie x;
        x.set(5);
        foo<pie>::bar(x);

        muffin y;
        y.set(5);
        //foo<muffin>::foo(y); //Causes a compilation Error because I only friended the pie specialization
}

Even notice where muffin friends the wrong foo, and still causes a compilation error. This works with both functions and classes. I am totally willing to accept that this isn’t possible in my specific situation (It’s actually looking more and more that way) I’d just like to understand why.

Answer 1

Befriending all possible specializations of baseclass2<D>::foo is rather easy:

template<class T> class baseclass1;

template<class D>
class baseclass2{
public:
  template<class T>
  void foo(D&, T&){ baseclass1<T> x; x.priv_foo(); }
};

template<class T>
class baseclass1{
  template<class D>
  template<class U>
  friend void baseclass2<D>::foo(D&, U&);

  void priv_foo(){}
};

template<class T>
class baseclass1{
  template<class D>
  template<class U>
  friend void baseclass2<D>::foo(D&, U&);
};

Live example.

A forward declaration of baseclass2 (so baseclass1 knows that baseclass2 exists and is a template) and two templates, one for the class, one for the function. It also looks like this for out-of-class definitions for function templates of class templates. 🙂

Befriending specifically baseclass2<D>::foo<T> is not possible, however, or I can’t find
the correct syntax for it.

A workaround might be some global function that forwards the access and together with the passkey pattern, but meh, it’s a mess (imho):

template<class D> class baseclass2;

template<class D, class T>
void baseclass2_foo(baseclass2<D>& b, D&, T&);

template<class D, class T>
class baseclass2_foo_key{
  baseclass2_foo_key(){} // private ctor
  friend void baseclass2_foo<>(baseclass2<D>&, D&, T&);
};

template<class T>
class baseclass1{
public: // public access, but only baseclass2_foo can create the key
  template<class D>
  void priv_foo(baseclass2_foo_key<D, T> const&){}
};

template<class D, class T>
void baseclass2_foo(baseclass2<D>&, D&, T&){
  baseclass1<T> x;
  x.priv_foo(baseclass2_foo_key<D, T>());
}

template<class D>
class baseclass2{
public:
  template<class T>
  void foo(D& d, T& t){ baseclass2_foo(*this, d, t); }
};

Live example.