Question

abstract class A {

    public void methodA() {
        System.out.println("methodA");
        methodB();
        showName();
    }

    public abstract void methodB();

    public void showName() {
        System.out.println("in showname base");
    }
}

class B extends A {

    public void methodB() {
        System.out.println("methodB");
    }

    public void showName() {
        System.out.println("in showname child");
    }
}

public class SampleClass {

    public static void main(String[] args) {
        A a = new B();
        a.methodA();
    }
}

Output is :

methodA
methodB
in showname child

Question :-

Since in overriding, the object type is considered. Is it the reason behind that class B’s showName() method called not of class A’s? If not then whats the cause of this output order?

Answer 1

It’s easy:

    A a = new B();
    a.methodA();

here is known that a is object of B class, so every method that can be overriden in class B is used from class B if there is no overriding, then method from class A must be used.

Considering order:

you invoke methodA that is declared as:

public void methodA() {
    System.out.println("methodA");
    methodB();
    showName();
}

from inside of methodA() you invoke both methodB() and showName(). They are overriden in class B, and object a is instanceof B, so that’s why they (from B class) are used.

EDIT as mentioned in comment:

@Jaikrat Singh, class B is still class A (child of it, but inheritance is relation of type: IS-A ). Class B has inherited methods from A class. So it has methodA as well. So it’s better to say, that methodA is also called from class B but with is default code – the same as provided in class A