Calling derived class method in base class - c++

Question

We can call "Base class methods" in derived class. but Can we call derived class method in base class function?
#include<iostream.h>
class base {
public:
void print()
{
cout<<"base class";
}
};
class derived : public base
{
public:
void print()
{
cout<<"derived class";
}
};
How to call this derived class print() function in Base class?

Answer 1

You cannot call a normal method defined in derived class from base class. What you can do is add a virtual (possibly pure) method in base class, provide implementation in derived class. You can call this method from base.
class Base{
public:
virtual void foo(){cout<<"Hello";};
void bar() { foo();}
};
class Derived: public Base{
public:
void foo() override{ cout<<"Hi";}
};
int main() {
Base* b1 = new Derived();
b1->bar(); //will call Derived::foo()
Base* b2=new Base();
b2->bar(); // will call Base::foo()
}

Answer 2

You can call derived class's method from base class method on a derived class object, if it's a virtual method:
class base {
public:
void doSomething() {
print();
}
virtual void print() {
cout<<"base class";
}
};
class derived : public base {
public:
virtual void print() {
cout<<"derived class";
}
};
int main() {
derived d;
base* pb = &d;
pb->doSomething();
}
It is known as Template method pattern.

Answer 3

If you know for certain that all instantiations of base will be as the base class of derived (as might be the case in a use of the Curiously Recurring Template Pattern), you can simply static_cast this to derived*:
#include <iostream>
class base {
public:
void call_derived_print();
void print()
{
std::cout<<"base class";
}
};
class derived : public base
{
public:
void print()
{
std::cout<<"derived class";
}
};
void base::call_derived_print() {
//undefined behaviour unless the most-derived type of `*this`
//is `derived` or is a subtype of `derived`.
static_cast<derived*>(this)->print();
}