C++: Implementing a function defined inside an anonymous structure

Question

Given the following code:

class Named {
  class /*Unnamed*/ {
    void Function();
  } un;
};

// Implement Named::Unnamed::Function here

int main() {
  Named named;
  named.un.Function();
}

Is there any way to implement Named::Unnamed::Function without either naming Unnamed or embedding the function's definition within the definition of Named?

I'm guessing the answer is "no", but GCC gives me the useful message "undefined reference to `Named::{unnamed type#2}::Function()', and it occured to me there might be some crazy possible syntax.

Answer 1

This is actually possible (in C++11), and in two ways:

struct Named {
  struct /*Unnamed*/ {
    void Function();
  } un;
  typedef decltype(un) Unnamed;
};

// #1
void Named::Unnamed::Function(){
}

//// #2
//typedef decltype(Named::un) Unnamed;
//void Unnamed::Function(){
//}
//// same way, using template alias
//template<class T> using alias = T;
//void alias<decltype(Unnamed::un)>::Function(){
//}

int main() {
  Named named;
  named.un.Function();
}

Live example on Ideone.

This is possible thanks to $9.1 [class.name] p5:

A typedef-name (7.1.3) that names a class type, or a cv-qualified version thereof, is also a class-name.