Overriding an equals method from the java…Object class

Question

I've been attempting to figure out the rationale behind this question and I have been struggling to see why the outcome is what it is. I will explain everything I understand and I hope someone will be able to fill in the blanks for me.

Imagine you have a class:

public class Point {
    public boolean equals(Object o) {
        if (o == null || (!(o instanceof Point)) { // Let's call this method 1
            return false;
        }
        Point other = (Point) o;
        return x == other.x && y == other.y;
    }

    public boolean equals(Point p) { // Let's call this method 2
        if (p == null) {
            return false;
        }
        return x == p.x && y == p.y;
    }
}

Now we create the following objects:

Object o = new Object()

Point p = new Point(3,4)

Object op = new Point(3,4)

If we call:

p.equals(o) // this calls method 1

p.equals(p) // this calls method 2

p.equals(op) // this calls method 1

However, this is where I get confused.

op.equals(o) // this calls method 1

op.equals(op) // this calls method 1

op.equals(p) // this calls method 1

Why does the last one call method 1? Shouldn't the method signature of method 2 warrant the call to go there?

If anyone could explain it to me that would be great!

Answer 1

op is a variable of type Object , which has no method having the signature public boolean equals(Point p) . Hence the only equals method that can be executed by calling op.equals() (regardless of the argument type) has the signature boolean equals (Object o) . Your Point class overrides boolean equals (Object o) , so your method 1 is called in all of the latter 3 cases.

Answer 2

Given that

Point p = new Point(3,4)
Object op = new Point(3,4)

and since op is an Object

op.equals(p)

will call the equals(Object o) method, because that's the only equals method that Object has.