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();
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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Last Updated :
28 Jun, 2021
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