CODEWITHSUNDEEP
javasethashset
jimao
jimao

Reputation: 55

Why HashSet doesn't build all the methods of the Set interface

I knew from my java courses that when i implement an interface i have to build all the methods that exist in that interface.But when i took a look of the HashSet class i observed that it didnt build all the methods of the Set interface.Why?

Upvotes: 0

Views: 236

Answers (3)

Paul Boddington
Paul Boddington

Reputation: 37655

HashSet implements all the methods of the Set interface. Not all of them appear directly in the HashSet code because some (e.g. removeAll) are inherited from AbstractSet, and others (e.g. toArray) are inherited from AbstractCollection. You can see all this in the documentation.

These abstract classes are extremely useful. If you extend AbstractSet, all you have to do is override size, contains and iterator, and you get correctly functioning versions of other methods (e.g. containsAll, hashCode, toArray, stream) for free. You may choose to override them anyway to make them more efficient.

Upvotes: 0

Jeff Morin
Jeff Morin

Reputation: 1010

If you look at the HashSet class declaration, you can see that besides implementing Set, Cloneable, and Serializable, it extends the AbstractSet class, which itself extends AbstractCollection.

For example, the two toArray() overloads are implemented in the AbstractCollection class. So are the addAll(), containsAll(), retainAll(), and removeAll() methods. Moreover, removeAll() is overridden in AbstractSet.

The equals() and hashCode() methods, which are defined in Object, are overridden in AbstractSet.

Hope this helps...

Jeff

Upvotes: 3

emsworth
emsworth

Reputation: 1171

Take a look at the superclass, AbstractSet. From the java docs: "This class provides a skeletal implementation of the Set interface to minimize the effort required to implement this interface." HashSet extends AbstractSet so anything not implemented in AbstractSet is implemented in HashSet.

Upvotes: 0

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