why we need to return 0 when Comparator.compare become equal

Question

I understand that when implement compare method of Comparator interface we need to return

• +1 if o1 > o2
• -1 if o1 < o2
• 0 if o1 == o2

my question is why we need to return 0 when both equal? what is the use case or where it being used? if we considered sorting when o2 greater than to o1 or o2 equal to o1 does not change it place. can anyone come explain practical use case for this?

Java documentation says

Compares its two arguments for order. Returns a negative integer, zero, or a positive integer as the first argument is less than, equal to, or greater than the second.

Does this mean return -1 or return 0 has same impact?

zero, or a positive integer

 @Override
    public int compare(Test f1, Test f2) {
        if (f1.getId() > f2.getId()) {
            return 1;
        } else if (f1.getId() < f2.getId()) {
            return -1;
        } else {
            return 0;
        }

    }

Answer 1

If you return -1 when the two values being compared are equal, compare(f1,f2) and compare(f2,f1) will both return -1. This means that the ordering of your elements will not be consistent. It can break some sorting algorithms.

That's why the general contract of compare requires that:

sign(compare(f1,f2)) = -sign(compare(f2,f1))

which means you must return 0 when the two values are equal.