从嵌套的 class 访问私有变量到包含 class 的私有方法

Question

//In file A.hpp
    class A
    {
        public:
        void random();
        A() {};
        virtual ~A() {};
        class B
        {
            public:
            B(int a, int b) 
              : c(a), d(b) {};
            virtual ~B() {};
            void SetVariable( int alpha )
            {
                beta = alpha;
            }
            private:
            int beta;
            int c; int d;
        };
        
        private:
        void GetVariable( int gamma ); 
    };
    
    int main()
    {
        
    }

I want to access beta in "GetVariable" method. beta is a private variable in nested class. I want to access the updated value of beta in GetVariable method.

Answer 1

You cannot. Even if the member was public you cannot access a member of one class in the member of a second class unless you have an instance of that second class available.

Your code looks a little complicated because it has inheritance and what not. Let me try to explain by means of a simpler example:

struct F { 
    struct G {
       int x; 
    };
   
    int get();
};

int main() {
    F::G g{42};
}

Now, similar question: How to return G::x from F::get() ?

You can't.

What you can do is...

Add a G member to F and return the member of that instance:

struct F { 
    struct G {
       int x; 
    };
    G g;  
    int get() {
         return g.x;
    }
};

or pass a G to get :

struct F { 
    struct G {
       int x; 
    };
    int get(const G& g) {
         return g.x;
    }
};

Just because G is defined inside F does not make any instance of G a member of a F .

---

Actually also the nesting is not that relevant because with regards to accessing the member G::x it is similar to, now with private :

 class G {
     int x;
 };

 struct F {
     int get() { /* how to access G::x here?!?! */ }
 };

Now again two issues: First you need an instance of G to access one of its members. Second, you cannot access it when it is private. As above, we pass an instance to get and in addition add an accessor:

 class G {
     int x;
 public:
     G() : x(42) {}
     int get() const { return x; }
 };

 struct F {
     int get(const G& g) { return g; }
 };

 int main() {
     G g;
     F f;
     int y = f.get(g);
 }

Alternatively you can declare F a friend of G , then G::get is not needed.

I suppose your code is a stripped down example of something more complex, because in the example code there is no reason to use F::get to get the value of G::x . Once you provided a way to access the member from outside the class, you can also access it directly in main without using F::get .