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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