模板类的朋友模板功能

Question

I have the following simplified code:

template <class T>
class A
{
public:
    template <class U>
    static U foo(T* p)
    {
        p;
        return U();
    }
};

class B
{
    /*template <class T>
    template <class U>
    friend U A<T>::foo<U>(T*);*/
    friend B A<B>::foo<B>(B*);
    B()
    {}
public:
};
...
A<B>::foo<B>(nullptr);

And it works quite well. But the things that I've not managed to do are commented:

/*template <class T>
template <class U>
friend U A<T>::foo<U>(T*);*/

I don't know what the syntax I should use to make it works. So I need to generalize my friend declaration to all possible types. I've tried quite a few variants of syntax but had no success. Could some one point me out what should I write instead of my commented code to make it works? Thanks!

Answer 1

What you're looking for is

template <class T>
template <class U>
friend U A<T>::foo(T*);

The following works on IdeOne.com

#include <iostream>

template <class T>
class A
{
public:
  template <class U>
  static U foo(T* p)
  {
    p;
    return U();
  }
};

class B
{
  template <class T>
  template <class U>
  friend U A<T>::foo(T*);

  B() {}

public:
  void hello() const
  {
    std::cout << "I'm a B!" << std::endl;
  }
};

int main(int, char*[])
{
  A<B>::foo<B>(NULL).hello();
}

Answer 2

I will agree with the othe commenters, friends and templates simply don't mix, at least not in any consistent way with across a range of compilers. The standard probably states exactly what should work, though that will likely not help you. Sad but true.

Friend is often not a good idea anyhow, it might be better to consider how to write the code without friendship, or use the "poor man's friend" of a public method with a comment indicating it is not for general use.