Virtual friend functions for a base class?

Question

I'm in the proccess of learning the language and this is a noob doubt.

Is it possible to use a virtual friend function? I don't know if it's possible, I didn't even test it but it could be useful in some situations. For example, for the overloaded operator<<().

DerivedClass dc;
BaseClass &rbc = dc;
cout << rbc;

My guess is it's possible, but I'm not sure since a friend function is not implemented in the class design, and theoretically is not part of it (though in this example, conceptually it makes sense that operator<<() should be a method, but due to syntax limitations it's not possible to implement it as one).

EDIT: my concern is related with this example:

BaseClass bc;
DerivedClass dc;
BaseClass *pArr[2];
pArr[1] = bc;
pArr[2] = dc;
for (int i = 0; i < 2; i++)
    cout << pArr[i];

in this array of mixed objects, I want the correct operator<<() called for each one.

Answer 1

Nope, friend virtual functions doesn't make sense at all.

friend functions are such, that are not methods (a.k.a. member functions) and have the right to access private/protected members of a class.

virtual functions can only be member functions. You can't have virtual non-member function.

---

You can make the operator<< take a reference to a base class and then call some virtual member function. This way, you can make the operator<< "almost virtual" :)

---

For example

class A
{
public:
    virtual void f() const { std::cout << "base"; }
};
class B: public A
{
public:
    virtual void f() const { std::cout << "derived"; }
};

std::ostream& operator<<(std::ostream& os, const A& a )
{
     a.f();
     return os;
}

int main()
{
    B b;
    std::cout << b << std::endl;

    return 0;
}

will print derived.