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.
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.
In C++, you can grant access beyond private on four levels.
public access (see pmr's answer)protected, irrelevant here)friend (see this answer)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:
...
};
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With