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Java | Inheritance | Question 1

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Output of following Java Program?
class Base {
    public void show() {
       System.out.println("Base::show() called");
    }
}
   
class Derived extends Base {
    public void show() {
       System.out.println("Derived::show() called");
    }
}
   
public class Main {
    public static void main(String[] args) {
        Base b = new Derived();;
        b.show();
    }
}

                    
(A) Derived::show() called (B) Base::show() called

Answer: (A)

Explanation: In the above program, b is a reference of Base type and refers to an object of Derived class. In Java, functions are virtual by default. So the run time polymorphism happens and derived fun() is called.

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Last Updated : 09 Jul, 2021
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