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.
Quiz of this Question
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