Virtual function in C++

Question

In the below c++ code using virtual functions

#include<iostream>
using namespace std;
class Base{
    public:
    virtual void fun(){
        cout << "Base::fun()called \n";
    }

};

class Child : public Base {
    public:
    void fun() {
        cout << "Child::fun() called\n";
    }

    void door(){
        cout << "Child::door() called \n";
    }
};


int main(){

    Base *base_ptr = new Child();
    base_ptr->fun();
    return 0;
}

How can I invoke door function using base_ptr? This question was asked in an interview. I was wondering whether it can be possible

Thanks for your replies

Answer 1

(Assuming that Base and Child cannot be modified.)

You can use static_cast to convert base_ptr to Child* .

static_cast<Child*>(base_ptr)->door()

This is safe as long as you are sure that base_ptr is actually pointing to a Child instance.

---

If you don't know what derived instance type base_ptr is pointing to, consider using dynamic_cast :

if(auto child = dynamic_cast<Child*>(base_ptr))
{
    child->door();
}

Unless the compiler manages to aggressively optimize it, dynamic_cast has extra runtime overhead compared to static_cast .

Answer 2

Because Base has no door() method, you obviously cannot simply call base_ptr->door() .

However, if you know base_ptr is a Child , you can cast it and then call door()

Child* child_ptr = static_cast<Child*>(base_ptr);
child_ptr->door();