Set by Default Sorted or Not?

UPDATED:

    Set s = new HashSet();
    s.add(1);
    s.add(5);
    s开发者_StackOverflow.add(4);
    s.add(9);
    s.add(7);
    s.add(8);        
    s.add("b");
    s.add("a");
    s.add("B");
    s.add("A");
    s.add("s");
    s.add("x");        
    s.add("d");        
    System.out.println(s);
    s.remove("b");
    s.remove("d");
    System.out.println(s);

Output:

[1, d, 4, b, 5, A, B, 7, a, 8, 9, s, x]
[1, 4, 5, A, B, 7, a, 8, 9, s, x]
[1, 4, 5, A, B, 7, a, 8, 9, s, x]

Need some information that Set default sort Integer value when we add but if i Add String to Set it would not sort by default.

Update: and Also Caps letter would always Sorted after runs many times.

java version "1.6.0_26" Java(TM) SE Runtime Environment (build 1.6.0_26-b03) Java HotSpot(TM) Client VM (build 20.1-b02, mixed mode, sharing)

please give me some idea. Thanks

HashSet does not guaranteed that its contents will be sorted in any way. There is a special interface for sets that do provide such a guarantee: it's called SortedSet:

A Set that further provides a total ordering on its elements. The elements are ordered using their natural ordering, or by a Comparator typically provided at sorted set creation time. The set's iterator will traverse the set in ascending element order. Several additional operations are provided to take advantage of the ordering. (This interface is the set analogue of SortedMap.)

In Java 6, there are two classes that implement this interface: ConcurrentSkipListSet and TreeSet.

No, HashSet is not sorted - or at least, not reliably. You may happen to get ordering in some situations, but you must not rely on it. For example, it's possible that it will always return the entries sorted by "hash code modulo some prime" - but it's not guaranteed, and it's almost certainly not useful anyway.

If you need a sorted set implementation, look at TreeSet.

From Oracle documentation we can find the HashSet class implements the set interface and internally backed by Hash Table.It makes no guarantees as to the iteration order of the set; in particular, it does not guarantee that the order will remain constant over time. But My code Which is given bellow finds it sorted even after running number of time.

 public static void main(String[] args){
        Random rand = new Random(47);           
        SortedSet<Integer> intset = new TreeSet<Integer>();
        Set<Integer> intsetUnsorted = new HashSet<Integer>();
        for(int i=0;i<10000;i++){
            intset.add(rand.nextInt(30));
        }
        intsetUnsorted.add(500);
        for(int i=0;i<10000;i++){
            intsetUnsorted.add(rand.nextInt(30));
        }
        String version = System.getProperty("java.version");
        System.out.println("JDK version-"+version);     
        System.out.println("TreeSet o/p");
        System.out.println(intset);
        System.out.println("Hash Set o/p");
        System.out.println(intsetUnsorted);

    }  

O/P

JDK version-1.6.0
TreeSet o/p
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29]
Hash Set o/p
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 16, 19, 18, 21, 20, 23, 22, 25, 24, 27, 26, 29, 28, 500]

I would suggest you check the Set interface javaDoc. In the All Known Implementing Classes section you have the provided implementations and if you check HashSet the first sentence answers your question.

Btw. checking javaDoc is always a good place to start ;-)