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If every object added to a java.util.HashSet implements Object.equals() and Object.hashCode() in a deterministic fashion, is the iteration order over the HashSet guaranteed to be identical for every identical set of elements added, irrespective of the order in which they were added?
Bonus question: what if the insertion order is identical as well?
(Assuming Sun JDK6 with same HashSet initialization.)
Edit: My original question was not clear. It is not about the general contract of HashSet, but what Sun's implementation of HashSet in JDK6 offers as guarantees concerning determinism. Is it inherently non-deterministic? What influences the order used by its Iterator?
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It means that HashSet does not maintains the order of its elements. Hence sorting of HashSet is not possible. However, the elements of the HashSet can be sorted indirectly by converting into List or TreeSet, but this will keep the elements in the target type instead of HashSet type.
There are three simple ways to iterate over a HashSet, which is the following : Using Iterator. Without using Iterator (using for loop) Using for-each loop.
No. HashSet is not ordered.
HashSet does not provide any method to maintain the insertion order. Comparatively, LinkedHashSet maintains the insertion order of the elements.
Absolutely not.
The insertion order directly influences the iteration order whenever you have a bucket collision:
When two elements end up in the same bucket, the first one that was inserted will also be the first one returned during iteration, at least if the implementation of collision handling and iteration is straightforward (and the one in Sun's java.util.HashMap is)
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