Child class inherit friendship

Question

I have a class State and another friend class Game. My question is, if I have a class MenuState which extends State, is MenuState a friend of Game as well?

EDIT:

So if I have a situation like this:

class Game
{
    private:
        StateManager* myStateManager = new StateManager();
}

class State
{
    public:
        static void create(StateManager* Parent, const std::string name) {};

    private:
        StateManager* parent;
}

#define DECLARE_STATE_CLASS(T)                                   \
static void create(StateManager* Parent, const std::string name) \
{                                                                \
        T* myState = new T();                                    \
        myState->parent = Parent;                                \
        Parent->manageState(name, myState);                      \
}                                                                \

class MenuState : State
{
    public:
        DECLARE_STATE_CLASS(MenuState)
}

int main()
{
    Game app;
    MenuState::create(This needs to be the stateManager in app, "MenuState");
    app.init("MenuState");
}

How can I make this work without making the state manager public?

No , check the following, friendship won't be inherited. http://stackoverflow.com/questions/3561648/why-does-c-not-allow-inherited-friendship — Zac Wrangler, Oct 03 '13 at 22:52

score 1 · Answer 1 · answered Oct 03 '13 at 22:53

1

So if the situation is like the following

class Game {
  friend class State;
}

class State {

}

class MenuState : public State {

}

Then the answer is no, as friend directive is not implicitly inherited. And I don't see why it should be inherited, Game explicitly allows State to access its internal implementation details but why this should be extended to subclasses of State which could do different things?

answered Oct 03 '13 at 22:53

Jack

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Alright, I have an object (declared private StateManager* myStateManager = new StateManager(); in Game) that I want MenuState to be able to access because I pass myStateManager to the MenuState::Create(StateManager* Parent, const std::string stateName) function. Without making myStateManager public, is this possible? – Oct 03 '13 at 22:59
I apologize for that. That was really confusing, not very good at formatting text in comments. If you don't quite understand what I mean or want me to post the exact code I'll edit the OP. – Oct 03 '13 at 23:00
Please see the original question. – Oct 03 '13 at 23:12

score 0 · Answer 2 · answered Oct 04 '13 at 04:44

You can add an access function to Game. Even if MenuState is a friend of Game, this still won't work, since you are trying to pass Game's private member, not access it inside the function.

If for instance, you have no right to modify Game, then the choice would be to add an access function inside State, and then you can use that access function from MenuState.

class State
{
    public:
    static void create(StateManager* Parent, const std::string name) {};
    StateManager* getManager(Game* g){
        return g->myStateManager;
    }

    private:
    StateManager* parent;
}

#define DECLARE_STATE_CLASS(T)                                   \
static void create(Game* g, const std::string name) \
{                                                                \
        T* myState = new T();                                    \
        myState->parent = State::getManager(g);                              \
        Parent->manageState(name, myState);                      \
}                                                                \

Child class inherit friendship

2 Answers2