C++ Is it possible that the base class has a function with a parameter of the derived class

Question

I want to do essentially the following:

class Base{
     void dosomestuff(Derived instanceOfDerived)
     {
         //Something
     }
};

class Derived : public Base{
     //Something
};

The Base needs a include of Derived, but to declare Derived it needs the declaration of Base first. Forward declaration does not work, because I do not want to use pointers.

Now my question: How do I accomplish that without pointers? Is that even possible? Or is the only possibility to make instanceOfDerived a pointer?

Answer 1

The original version of the question asks about having Base hold Derived as a data member. That's obviously impossible. It's the same infinite recursion problem (a Derived contains a Base subobject, so a Base would recursively contain itself, and it will be turtles all the way down.)

The revised question asks about having a member function of Base taking a by-value argument of type Derived. That's perfectly possible. You need a forward declaration of Derived, and to defer the definition of the member function until after the actual definition of Derived:

class Derived;
class Base{
     void dosomestuff(Derived instanceOfDerived); // declaration only
};

class Derived : public Base{
     //Something
};

// out-of-class definition, so making it explicitly inline
// to match the in-class definition semantics
inline void Base::dosomestuff(Derived instanceOfDerived) {
     //Something
}

Demo.