C++ | Virtual Functions | Question 2

Predict output of the following program

#include<iostream>
using namespace std;

class Base
{
public:
    virtual void show() { cout<<" In Base \n"; }
};

class Derived: public Base
{
public:
    void show() { cout<<"In Derived \n"; }
};

int main(void)
{
    Base *bp = new Derived;
    bp->show();

    Base &br = *bp;
    br.show();

    return 0;
}

(A)

In Base 
In Base 

(B)

In Base 
In Derived

(C)

In Derived
In Derived

(D)

In Derived
In Base 

Answer: (C)

Explanation: Since show() is virtual in base class, it is called according to the type of object being referred or pointed, rather than the type of pointer or reference.

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