Consequence when compareTo() is inconsistent with equals()

Question

Can somebody put some light on what are the Consequences when compareTo() is inconsistent with equals() of a class. I have read that if Obj1.compareTo(Obj2) = 0 then it's not mandatory to be Obj1.equals(Obj2) = true. But what is the consequence if this happens. Thanks.

Answer 1

The documentation for Comparable explains this in some detail:

The natural ordering for a class C is said to be consistent with equals if and only if e1.compareTo(e2) == 0 has the same boolean value as e1.equals(e2) for every e1 and e2 of class C. Note that null is not an instance of any class, and e.compareTo(null) should throw a NullPointerException even though e.equals(null) returns false.

It is strongly recommended (though not required) that natural orderings be consistent with equals. This is so because sorted sets (and sorted maps) without explicit comparators behave "strangely" when they are used with elements (or keys) whose natural ordering is inconsistent with equals. In particular, such a sorted set (or sorted map) violates the general contract for set (or map), which is defined in terms of the equals method.

For example, if one adds two keys a and b such that (!a.equals(b) && a.compareTo(b) == 0) to a sorted set that does not use an explicit comparator, the second add operation returns false (and the size of the sorted set does not increase) because a and b are equivalent from the sorted set's perspective.

Virtually all Java core classes that implement Comparable have natural orderings that are consistent with equals. One exception is java.math.BigDecimal, whose natural ordering equates BigDecimal objects with equal values and different precisions (such as 4.0 and 4.00).