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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