How the key word super works in java-Java Puzzle

Question

public class B {

    public B() {

    }

    private void m0(){
        System.out.println("BO");
    }
    public void m1(){
        System.out.println("B1");

    }

    public void test(){
        this.m0();
        this.m1();
    }
}



public class D extends B{

    /**
     * 
     */
    public D() {

    }

    public void m0(){
        System.out.println("DO");
    }
    public void m1(){
        System.out.println("D1");

    }

    public void test(){
      super.test();
    }

    public static void main(String[] args) {
        B d=new D();
        d.test();
    }


}

My question is why the output is BO,D1 instead of BO,B1. I am not getting how the super keyword plays the role of calling the methods of the child class instead of the parent class.

Answer 1

Because your method m0 in class B is private, it is not overridden by class D.