how does Set understands duplicate elements?

[amit daundkar]

How does Set understand if duplicate element is inserted? I mean does it perform "equals" test everytime an object is inserted?

Please reply.
Thanks.

[Viktor Kubinec]

In order to understand this, you should first understand hashCode and equals contract.

Find HashSet on this site :
Set interface

If hashCodes equal and objects equal, then the insertion is resolved as duplicated element and the add method returns false.

[Henry Wong]

It''s also implementation specific. For example, the TreeSet uses either the Comparable or Comparator interfaces.

Henry

[amit daundkar]

Thank you Viktor, Henry.

[Hemant B. Kumar]

Henry Wong wrote:
It''s also implementation specific. For example, the TreeSet uses either the Comparable or Comparator interfaces.

Yeah that's right but I think in that case also it would be using equals and hashcode method, isn't it?

[Henry Wong]

Hemant B. Kumar wrote:

Henry Wong wrote:
It''s also implementation specific. For example, the TreeSet uses either the Comparable or Comparator interfaces.

Yeah that's right but I think in that case also it would be using equals and hashcode method, isn't it?

Interestingly, the answer is no. The reason it is interesting is because if you look at the JavaDoc for java.util.Set, it is actually defined as using the equals() method to determine duplicates. However, the java.util.TreeSet class does not use the method -- and the JavaDoc explicitly mentions that it is the programmers responsibility to guarantee that it is consistent -- meaning that if Comparable or Comparator says that two objects are equal, the equal() method should return equal.

Henry

[Rob Spoor]

If it isn't you can get odd behaviour. For instance:

Set<String> set1 = new HashSet<String>();
Set<String> set2 = new TreeSet<String>(); // natural ordering
Set<String> set3 = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER); // case insensitive natural ordering

set1.add("Hello World");
set2.add("Hello World");
set3.add("hello world"); // different case

System.out.println(set1.equals(set2)); // true as expected
System.out.println(set1.equals(set3)); // false as expected
System.out.println(set2.equals(set1)); // true as expected
System.out.println(set2.equals(set3)); // false as expected
System.out.println(set3.equals(set1)); // true
System.out.println(set3.equals(set2)); // true

What you now get is that set3.equals(set1) but !set1.equals(set3).