Doubt about the concept of HashSet and LinkedHashSet

Question

I read one of the SCJP 6 exam book, when talking about Set, it gives one definition:
When using HashSet or LinkedHashSet, the objects you add to them must override hashCode(). If they don't override hashCode(), the default Object. hashCode() method will allow multiple objects that you might consider "meaningfully equal" to be added to your "no duplicates allowed" set.
What I am getting confused is that IF the objects we add to them must override hashCode(), we must override equals() method also! Isn't it?
Edited by: roamer on 2009?10?23? ??10:29

Answer 1

jverd wrote:
endasil wrote:
When using HashSet or LinkedHashSet, the objects you add to them must override hashCode(). If they don't override hashCode(), the default Object. hashCode() method will allow multiple objects that you might consider "meaningfully equal" to be added to your "no duplicates allowed" set.This really is completely wrong. Duplicates being added to your set has nothing to do with not overriding hashCode, and everything to do with not overriding equals. No, if you override equals but not hashCode, you can get dupes. That is, two items that your equals method says are equal can make it into the Set.Sorry, to clarify, what I meant by that was that to avoid duplicates of meaningfully equal objects, you must override equals, and overriding hashCode is just a consequent of overriding equals (to maintain the equal objects have equal hashcodes invariant). Overriding hashCode alone will not, and cannot, prevent meaningfully equal duplicates if you don't first define meaningfully equal by overriding equals.
In summary, it's wrong because
1) You don't have to override hashCode() if you don't override equals (or rather, don't deviate from the default "two objects are equal if and only if they are the same object").
2) If you have any other definition of meaningfully equal that you want enforced in a Set, you must override equals to correspond to that definition.
3) If you override equals, you must override hashCode by the contract of hashCode and equals.
That's why I was saying that the statement was completely wrong. It's taking a backwards approach.
By not overriding equals, you're saying that no two separate instances can be meaningfully equal, so the default hashCode is fine. Yeah, I kind of figured for the statement to be meaningful, it is assumed that you have already overridden equals. Otherwise what's the point of even mentioning something two objects "meaningfully equal"?All this talk of "meaningfully equal" makes me think the opposite: that they haven't overridden equals, or haven't discussed it yet. Otherwise you could just say that the objects are equal, or are equal according to equals().
I would question the authority of the source that said that.I would quesiton the source's ability to express himself clearly. :-)Yeah, maybe that's all it is. In that case, I suggest (to the OP) reading [Effective Java, Chapter 3|http://java.sun.com/developer/Books/effectivejava/Chapter3.pdf]. Best source I've seen for beginners to make sense of all this.
Edited by: endasil on 26-Oct-2009 1:11 PM