“elaborated type refers to typedef” error when trying to befriend a typedef

Question

Let\'s say I have the following piece of code (a simple CRTP class hierarchy). I want to typedef the base class type to save myself typing (in my actual code, I use the base

Answer 1

Enable C++11 only and use friend BaseType

You cannot use friend class on typedef in C++03.

An elaborated-type-specifier shall be used in a friend declaration for a class(101)

101) The class-key of the elaborated-type-specifier is required.

elaborated-type-specifier:

class-key ::opt nested-name-specifieropt identifier

class-key ::opt nested-name-specifieropt templateopt template-id

enum ::opt nested-name-specifieropt identifier

typename ::opt nested-name-specifier identifier

typename ::opt nested-name-specifier templateopt template-id

Answer 2

I believe that this is not possible with C++03, but was added to C++11 in which you can simply omit the class keyword:

friend BaseType;

Answer 3

You actually can do it in C++ earlier than C++11, but it requires a fairly elaborate (haha) workaround, along the lines of, 1st declare this "helper":

template< class T > struct ParamTypeStruct { typedef T ParamType; };

Then, your friend declaration:

friend class ParamTypeStruct< BaseType >::ParamType;