Re: how to make nested class in a class template a friend of some other class template?

From:
Greg Herlihy <greghe@pacbell.net>
Newsgroups:
comp.lang.c++.moderated
Date:
Mon, 20 Aug 2007 04:24:02 CST
Message-ID:
<C2EE2F9D.798%greghe@pacbell.net>
On 8/19/07 4:06 PM, in article
1187557675.894924.130450@k79g2000hse.googlegroups.com, "seaswar"
<drhookeypookey@gmail.com> wrote:

In the following, how do I make class C<T>::N a friend of template
class X<U>. I tried various combinations. None worked. Complains that
X::fx is private. One even caused g++ to dump core.
Thanks
Suresh

template<typename T>
class C;

template<typename U>
class X {
private:
    static void fx() { }

    //I tried the following which did not work:
    template<typename T>
    friend class C;

    //Neither did this:
    template<typename T>
    friend class C<T>::N;

    //The following dumped core (gcc3.4.4 on cygwin):
    template<typename T>
    friend void C<T>::N::fn();
};


The friend declaration in X should specify which C template specialization
contains the friend class N. Presumably, the friend N class is to be found
in the C template specialization that has the same template type parameter
as the X specialization itself:

     template< typename T>
     class C;

     template< typename U>
     class X
     {
     private:
         static void fx() { }

         friend class C<U>::N;
     };

template<typename T>
class C {
public:
    class N {
    public:
        void fn()
        {
            X::fx(); // problematic call
        }
    };
};


The call to fx() within N::fn() needs to specify to which X specialization
the fx() static method belongs. Since the friend declaration in X makes
C<T>::N a friend of X<T> only, then X<T>::fx() is the only X specialization
whose fx() method is accessible in the current context:

     template< typename T>
     class C
     {
     public:
         class N
         {
         public:
             void fn()
             {
                 X<T>::fx(); // only X<T> is C<T>::N's friend
             }
         };
     };

Greg

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