Java | Inheritance | Question 1
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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