Comparing objects in HashSet

Question

I'm making a 2D game that has a stars in it. I decided to create constructor in class named Star that gives random coordinates.

public Star(){
    super(0,0);
    x = randomX.nextInt(maxX - minX + 1);
    y = randomY.nextInt(maxX - minY + 1);
}

Then, in other class I put them in HashSet

Set<Star> star = new HashSet<>();

public Set<Star> generateStars(){
    while (star.size() < numberOfStars){
            star.add(new Star());
    }
    return star;
}

Of course, I have render and tick methods but I think it's not worth to paste them. My lecturer told me that there can be same stars and to prevent that I should use identity function using hashcodes. Can someone help me figure that out ? I imagine that this function should check if the hashcodes are the same and if it's the case it should return only one value that way we will add 1 object instead of 2 into the HashSet. Am I right ?

Answer 1

Overriding the hashCode() method alone in your Star class will not work you will have to override the equals() method.

See the following code where we don't override the equals() method:

class Star {
    int x, y;

    public Star(int x, int y) {
        this.x = x;
        this.y = y;
    }

    @Override
    public int hashCode() {
        return Objects.hash(x, y);
    }
}

public class Main {
    public static void main(String[] args) {
        Star s1 = new Star(0, 0);
        Star s3 = new Star(0, 0);
        Star s2 = new Star(31, -31*31);
        Set<Star> set = new HashSet<>();
        set.add(s1);
        set.add(s2);
        System.out.println(set.size());
    }
}

This will print 3 (instead of 2 what you might expect).

The reason for this is that the add method of java.util.Set compares 2 objects based on equals() method and not based on hashCode() method.

In the above code for the Star class, if you add the equals() method the output will be 2 now. For your reference the you can override the equals() method as follows:

@Override
public boolean equals(Object startObject) {
    if (this == startObject) return true;
    if (startObject == null || getClass() != startObject.getClass()) return false;
    Star star = (Star) startObject;
    return x == star.x &&
            y == star.y;
}

So why do you need to add hashCode()?

1. As you are using HashSet the add method behind the scene will call the equals() method and will also call hashCode() to decide the bucket in which the new object should be put. To maintain the contract of hashCode() and equals() Both should be overridden.
2. When ever you override equals(), it is recommended to override hashCode() also. (Vice-versa is also true). See this link for details.

Contract for hashCode() and equals(): If for two objects say o1 and o2, o1.equals(o2) is true then hash of o1 and o2 should be same.

Make sure that you understand this properly, from the above statement it is not implied that if the hash of 2 objects are same, then o1.equals(o2) should return true. It can be possible that for 2 objects, their hash is same but the o1.equals(o2) returns false

See here, what Object's hashCode() method guarantees.

See this link to get more detailed information about this topic.