Why can't I declare a friend through a typedef?

Question

Does anyone know why typedefs of class names don\'t work like class names for the friend declaration?

class A
{
public:
};

class B : public A
{
public:
   t         


        

Answer 1

It can't, currently. I don't know the reason yet (just looking it up, because i find it interesting). Update: you can find the reason in the first proposal to support typedef-names as friends: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2003/n1520.pdf . The reason is that the Standard only supported elaborated-type-specifiers. It's easy to allow only those, and say if the entity declared as friend is not declared yet, it will be made a member of the surrounding namespace. But this means that if you want to use a template parameter, you would have to do (a class is required then for example)

friend class T;

But that brought additional problems, and it was figured not worth the gain. Now, the paper proposes to allow additional type specifiers to be given (so that this then allows use of template parameters and typedef-names).

The next C++ version (due to 2010) will be able to do it.

See this updated proposal to the standard: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2005/n1791.pdf . It will not allow only typedef names, but also template parameters to be used as the type declared as friend.