成为模板类的另一个实例的朋友

Question

template <typename T> class Foo;    
template <typename T> int g(Foo<T> const&);

template <typename T> class Foo
{
public:
    template <typename U> int f(Foo<U> const& p) const { return p.m; }

    // which friend declaration will allow the above function to compile? The
    // next one doesn't work.
    template <typename U> friend void Foo<U>::template f<T>(Foo<T> const&) const;

    // while this one work for g().
    friend int g<T>(Foo<T> const&);

private:
    int m;
};

template <typename T> int g(Foo<T> const& p) { return p.m; }

// Let's call them
void bar()
{
    Foo<int> fi;
    Foo<double> fd;
    fd.f(fi);
    g(fi);
}

The above doesn't compile with g++ nor Como. g() is here to show what I would like to do with f().

For instance, here are g++ messages:

foo.cpp:11: error: invalid use of template-id ‘f<T>’ in declaration of primary template
foo.cpp: In member function ‘int Foo<T>::f(const Foo<U>&) const [with U = int, T = double]’:
foo.cpp:27:   instantiated from here
foo.cpp:17: error: ‘int Foo<int>::m’ is private
foo.cpp:7: error: within this context

and como's one:

"ComeauTest.c", line 11: error: an explicit template argument list is not allowed on
          this declaration
      template <typename U> friend void Foo<U>::template f<T>(Foo<T> const&) const;
                                        ^

"ComeauTest.c", line 7: error: member "Foo<T>::m [with T=int]" (declared at line 17)
          is inaccessible
      template <typename U> int f(Foo<U> const& p) const { return p.m; }
                                                                    ^
          detected during instantiation of "int Foo<T>::f(const Foo<U> &)
                    const [with T=double, U=int]" at line 27

2 errors detected in the compilation of "ComeauTest.c".

Variants suggested by the error messages didn't either.

BTW, I know of the obvious work around

template <typename U> friend class Foo<U>;

Edit:

14.5.4/5 (of n3225, 14.5.3/6 of C++98 is similar but the following text is clearer in n3225) starts by

A member of a class template may be declared friend of a non-template class...

which could imply that a member of a class template may not be declared friend of a template class but my first interpretation would that this sentence was an introduction for the following explanations (mainly they apply to any specialisation, explicit or not, given that the prototype is correct).

Answer 1

I think you have to say template <> before the friend declaration so it knows you're friending a specialization.

EDIT: I can't get any error with your code, even making an instantiation and calling g . Can you post a minimal set of actual code that's causing the error, along with the error message?

Answer 2

Though I'm not 100% sure this is strictly standard conforming, the following code can be compiled by ideone(gcc-4.3.4) and Comeau online:

template< class >
class Foo {
  int m;
 public:
  template< class U > int f( Foo<U> const& p ) const { return p.m; }
  template< class U > template< class V >
  friend int Foo<U>::f( Foo<V> const& ) const;
};

void bar() {
  Foo<int> fi;
  Foo<double> fd;
  fd.f( fi );
}

Hope this helps.