C++ Templates: Partial Template Specifications and Friend Classes

Question

is it possible to somehow make a partial template specification a friend class? I.e. consider you have the following template class

template <class T> class X{
    T t;
};

Now you have partial specializations, for example, for pointers

template <class T> class X<T*>{
    T* t;
};

What I want to accomplish is that every possible X<T*> is a friend class of X<S> for ANY S. I.e. X<A*> should be a friend of X<B>.

Of course, I thought about a usual template friend declaration in X:

template <class T> class X{
    template <class S> friend class X<S*>;
}

However, this does not compile, g++ tells me this:

test4.cpp:34:15: error: specialization of 'template<class T> class X' must appear at namespace scope

test4.cpp:34:21: error: partial specialization 'X<S*>' declared 'friend'

Is this not possible at all or is there some workaround?

The reason why I am asking is that I need a constructor in X<T*> that creates this class from an arbitrary X<S> (S must be a subtype of T).

The code looks like this:

template <class T> class X<T*>{
    T* t;

    template<class S>
    X(X<S> x) : t(&(x.t))  {} //Error, x.t is private
}

Now, the compiler complains, of course, that x.t is not visibile in the constructor since it is private. This is why I need a partial specialization friend class.

Answer 1

In C++, you can grant access beyond private on four levels.

• completely public access (see pmr's answer)
• access within inheritance hierarchy (protected, irrelevant here)
• to a base template friend (see this answer)
• to a non-template or fully specialized friend (too weak to solve your use case)

There is no middle way between the two latter kinds of friendship.

From §14.5.4 of the C++ standard:.

Friend declarations shall not declare partial specializations.

The following declaration will allow you to implement what you need. It gives you a free hand to access any specialization of your template from any other specialization, but still only within X. It is slightly more permissive than what you asked for.

template<class T> class X
{
    template<class Any> friend class X;
    public:
        ...
};

Answer 2

We can define a getter protected by a key defined in X.

#include <type_traits>

template <class T> class X{
  T t;
public:
  struct Key {
    template<typename S>
    Key(const X<S>&) {
      static_assert(std::is_pointer<S>::value, "Not a pointer");
    }
  };

  const T& get(Key) const { return t; }
  T& get(Key) { return t; }
};

template <class T> class X<T*> {
  T* t;
public:
  template<class S>
  X(X<S>& x) : t(&(x.get(typename X<S>::Key(*this))))  {}
};

int main()
{
  X<int> x1;
  X<int*> x2(x1);
  return 0;
}

This still has some weakness. Everybody with an X<T*> can now use get. But this is so obfuscated by now, that no one is goiing to realize that. I'd choose a simple public getter.