Method overriding and overloading

Question

I have the code listed below. Why does it print: "V" - "greet(Z)" - "greet(Z)"? I would have said "V" - "greet(V)" - "greet(Z)" but it seems that I'm missing some points on method overloading and overriding, can someone explain me and/or link some resources to master this?

class Z {

    public void me() {
        System.out.print(" Z");

    }

    public void greet(Z z) {
        System.out.print ("greet(Z)");
    }

}
class V extends Z {

    @Override
    public void me() {
        System.out.println("V");
    }

    public void greet(V v) {
        System.out.println("greet(V)");
    }
}
public  class Quiz{
    public  static void main(String[] args) {

        Z a = new V();
        V b = new V();
        a.me();
        System.out.print("-");
        a.greet(b);
        System.out.print("-");
        a.greet(a);
    }
}

Answer 1

public void greet(V v) is an overloaded version of public void greet(Z z) : the former does not override the latter.

a.greet(b);

a is of type Z , there is only one method to choose from: greet(Z z) .

a.greet(a);

a is of type Z , there is still only one method to choose from: greet(Z z) .

b.greet(a);

b is of type V , there are two methods to choose from. Since a is of type Z , the most appropriate greet(Z z) will be chosen.

b.greet(b);

b is of type V , there are two methods to choose from. Since b is of type V , the most appropriate greet(VV) will be chosen.

Answer 2

When you initialize

  V b = new V();

b has a reference of V and object is created of type V so when u call

 a.greet(b);

It will first look into V class that is accepting parameter of type V.

But in the first line

Z a = new V();

You are creating object of V but the type is Z.So when calling

a.greet(b);

will only accept the type Z. Because the the greet method in V is overloaded to accept type V.