Question

equals and hashCode method must be consistent, which means that when two objects are equal according to equals method their hashCode method should return the same hash value.

Java returns a unique hash code if we do not override the hashCode() method.

class HashValue {

    int x;

    public boolean equals(Object oo) {
        // if(oo instanceof Hashvalue) uncommenting ths gives error.dunno why?
        // :|
        HashValue hh = (HashValue) oo;

        if (this.x == hh.x)
            return true;
        else
            return false;
    }

    HashValue() {
        x = 11;
    }

}

class Hashing {
    public static void main(String args[]) {
        HashValue hv = new HashValue();
        HashValue hv2 = new HashValue();

        System.out.println(hv.hashCode());
        System.out.println(hv2.hashCode());

        if (hv.equals(hv2))
            System.out.println("EQUAL");
        else
            System.out.println("NOT EQUAL");
    }
}

Why does uncommenting the line gives compilation error?

If the objects have unequal hash codes, why are they shown equal even though the default hashcode varies?

Answer 1

Equality is only determined by method equals(). And method hashCode() is used in other situations, like by Map or Set. It is somewhat like a pre-condition or hint before actually calling equals (for efficiency). So it is assumed that if 2 objects are equal (that is, equals() returns true), then their hashCodes() must return the same value.

So in your code, 2 objects are equal, as long as your overriden equals() returns true, no matter what hashCode() does. hashCode() is not called at all when comparing for equality.

This question has more in-depth information regarding to the relationship between equals() and hashCode().