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C++ | Virtual Functions | Question 14

  • Difficulty Level : Medium
  • Last Updated : 28 Jun, 2021

Predict the output of following C++ program

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