C++ | Virtual Functions | Question 14

Predict the output of following C++ 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->Base::show();  // Note the use of scope resolution here
    return 0;
}

(A) In Base
(B) In Derived
(C) Compiler Error
(D) Runtime Error


Answer: (A)

Explanation: A base class function can be accessed with scope resolution operator even if the function is virtual.

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