friendship from derived class method to base class members

Patrik Published at Dev

Patrik

I would like to know if there's a way to make a method from a derived class a friend of its base class. Something like:

class Derived;
class Base
{
    int i, j;
    friend void Derived::f();
protected:
    Base();
};

class Derived : public Base
{
public:
    void f();
};

The errors I got were:

error: C2027: use of undefined type 'Derived'
see declaration of 'Derived'
error: C2248: 'Base::i' : cannot access private member declared in class 'Base'
see declaration of 'Base::i'
see declaration of 'Base'
error: C2248: 'Base::j' : cannot access private member declared in class 'Base'
see declaration of 'Base::j'
see declaration of 'Base'
error: C2027: use of undefined type 'Derived'
see declaration of 'Derived'

I struggled with it during all the day. Everything I found about friendship use only separated classes, not inheritance.

quantdev

No there is no direct way : Base needs the definition of Derived::f while Derived also needs the definition of it's Base class.

But it does not matter, you should not do that, you can, in order of preference :

Provide protected accessors in the Base class
Make the entire Derived class a friend (not necessary in general)
You can use an intermediate helper class which only forward the call of this specific method, and give it friendship :

Example here:

class Base;
class Derived;

class Helper final
{
    friend class Derived;

    public:
        void f(Base* base);
    private:
        Helper() {}
};

class Base
{
    int i, j;
    friend class Helper;
protected:
    Base() {}
};

class Derived : public Base
{
public:
    void f();
private:
    Helper helper;
};

void Helper::f(Base* base)
{
    base->i = 10; base->j = 5;
    std::cout << "Help !" ;
}

void Derived::f()
{
    helper.f(this);
}

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