#include<iostream> using namespace std;
class Base {
public :
void fun() { cout << "Base::fun() called" ; }
void fun( int i) { cout << "Base::fun(int i) called" ; }
}; class Derived: public Base {
public :
void fun() { cout << "Derived::fun() called" ; }
}; int main() {
Derived d;
d.Base::fun(5);
return 0;
} |
(A) Compiler Error
(B) Base::fun(int i) called
Answer: (B)
Explanation: We can access base class functions using scope resolution operator even if they are made hidden by a derived class function.
Quiz of this Question