Why need to be a friend to access protected members?

Question

When I'm in Sub::f() and try to access the protected members of another sub-class to which the Base* b pointer points to, it won't compile until Sub is a friend of Base. Why do I need to do this?

class Base{
//friend class Sub;
protected:
    int i;
    virtual void f() = 0;
};

class Sub : public Base{
    Base* b;
public:
    Sub(Base* ba) : b(ba){}
    void f(){
        b->f();
        cout << b->i << endl;
    }
};

because you can only lets your friends touch your private parts. — Scary Wombat, Jan 29 '14 at 08:05
But `Sub` inherits everything from `Base`. Don't derived classes have access to protected members of their base-class? — Daniel, Jan 29 '14 at 08:08
if you are having a specific issue try posting a more complete and self contained example. — user2485710, Jan 29 '14 at 08:14
@Daniel _'But Sub inherits everything from Base.'_ So why do you have an extra member for `Base`?? — πάντα ῥεῖ, Jan 29 '14 at 08:15
Ok I guess `Sub` has to be a friend because `Base* b` has another object and it's unimportant whether the access happens in a sub-class or elsewhere -> the accessing object/function has to be a friend. — Daniel, Jan 29 '14 at 08:21
@πάνταῥεῖ I have something like a composite-pattern-structure (one sub-class can have another sub-class or is a endpoint). — Daniel, Jan 29 '14 at 08:23

score 1 · Answer 1 · answered Jan 29 '14 at 08:07

1

protected can be accessed only in hierarchy, but if you want exception, making friend is the way.

Friend class can access both private and protected members of the class.

answered Jan 29 '14 at 08:07

Pranit Kothari

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score 0 · Accepted Answer · answered Jan 29 '14 at 08:44

protected members are accessible in own class and in derived class but not outside of the class so if you want to do it, use 'friend'. And because of this feature provided by CPP,CPP is not 'fully or complete' object oriented programming language (loophole) as we can access private members too with friend's help.

score 0 · Answer 3 · answered Jan 29 '14 at 08:47

0

b->f() and b->i are not accessing f and b of his own base, but of somebody else base (the one pointed by b)

sub    base
|____b_|___i___|
     |
     |_
       |
       V
sub    base
|____b_|___i___|

answered Jan 29 '14 at 08:47

Emilio Garavaglia

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Thx;) Thats a nice picture. – Daniel Jan 29 '14 at 13:55

Why need to be a friend to access protected members?

3 Answers3