How to access protected members in a derived class?

Question

From http://www.parashift.com/c++-faq-lite/basics-of-inheritance.html#faq-19.5

A member (either data member or member function) declared in a protecte

Answer 1

You are not accessing protected function in your derived class, you are trying to overload it and promote from protected to public. This is a forbidden action, you only can hide functions in derived class, e.g. overload protected function as a private.

Accessing protected function means call it from some member of a class:

class Y : public X
{
public:
    void call() {
      fun();
    }
}

or like you call it by objX.fun(); is also correct.

Answer 2

One solution is to add friend class Y to X.

Answer 3

In void Y::call ()

    X objX;
    objX.fun (); 

// here you're trying to access the protected member of objX , this/current object of Y doesn't contains objX as it's base object , they both are different objects. That is why you can't access its member.

you can make Y friend of X as a solution told by @anycom.

Edit: what I mean is, from inside Y (which is inherited from X) , you can simply call protected/public members of "it's" base class X i.e you're accessing it's base members simply. But That doesn't means you can now access the protected members of all objects of type X, since you're trying to access those members from outer scope of class X i.e via object of X. You know all these rules but it seem you did too much thinking 0_o

Answer 4

Well, if friend is ok, then this angle may as well be ok:

#include <iostream>

class X {
private:
    int var;
protected:
    virtual void fun() {
        var = 10;
        std::cout << "\nFrom X" << var;
    }

    static void Fun(X& x) {
        x.fun();
    }
};

class Y : public X {
private:
    int var;
public:
    virtual void fun() {
        var = 20;
        std::cout << "\nFrom Y" << var;
    }

    void call() {
        fun();
        X objX;
        objX.fun(); /* << ne-ne */
        Fun(objX); /* << ok */
    }
};

Of course, be mindful of the type you pass to X::Fun if you use this as-is.

Answer 5

See this example:

#include <iostream>
using namespace std;

class X {
  private:
    int var;
  protected:
    void fun () 
    {
      var = 10;
      cout << "\nFrom X" << var; 
    }
};

class Y : public X {
  private:
    int var;
  public:
    void fun () 
    {
      var = 20;
      cout << "\nFrom Y" << var;
    }

    void call ()
    {
      fun(); /* call to Y::fun() */
      X::fun (); /* call to X::fun() */

      X objX;
      /* this will not compile, because fun is protected in X        
      objX.fun (); */

    }
};

int main(int argc, char ** argv)  {
  Y y;
  y.call();
  return 0;  
}

This yields

From Y20
From X10

Because you have overloaded the fun()-method in Y, you have to give the compiler a hint which one you mean if you want to call the fun-method in X by calling X::fun().

Answer 6

I think that the thing you are trying to do should looks like this:

#include <iostream>
using namespace std;

class X
{
    private:
        int var;
protected:
    virtual void fun () 
    {
        var = 10;
        cout << "\nFrom X" << var; 
    }
};

class Y : public X
{
private:
    int var;
public:
    virtual void fun () 
    {
        var = 20;
        cout << "\nFrom Y" << var;
    }

    void call ()
    {
        fun ();


        X::fun ();
    }
};

That way you can invoke hiden member from your base class. Otherwise you have to add friend X as it was pointed in other post.