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up vote 0 down vote favorite

I am trying to retrieve random elements from a linkedhashset. Below is my code but it is giving me exception everytime.

private static void generateRandomUserId(Set<String> userIdsSet) {

    Random rand = new Random(System.currentTimeMillis());
    String[] setArray = (String[]) userIdsSet.toArray();
    for (int i = 0; i < 10; ++i) {
        System.out.println(setArray[rand.nextInt(userIdsSet.size())]);
    }
}

Below is the exception I am getting-

java.lang.ClassCastException: [Ljava.lang.Object; incompatible with [Ljava.lang.String;

Can anyone help me on this? And is there any better way of doing this?

share|improve this question
    
You should convert your array using userIdsSet.toArray(new String[]) to generate a String array. Otherwise the method you are using will generate an Object array. – Edwin Dalorzo Jun 19 '13 at 21:33
up vote 11 down vote accepted

Try this:

String[] setArray = userIdsSet.toArray(new String[userIdsSet.size()]);

the toArray method that takes no arguments returns an Object[] which can't be cast to a String[]. The other version returns a typed array.

share|improve this answer
    
Perhaps explain why? – T.J. Crowder Jun 19 '13 at 21:32
    
@assylias Good point! In the link you provided they use new String[0]... I tested it and it works as well as your example – Paolof76 Jun 19 '13 at 21:50
1  
@Paolof76 using new String[0] allocates an unnecessary object so I prefer the version in my answer where possible (although the difference is minimal in most situations). – assylias Jun 19 '13 at 21:51
    
@assylias why it is different from new String[userIdsSet.size()]? – Paolof76 Jun 19 '13 at 21:54
2  
@Paolof76 if you use set.toArray(new String[0]), you create an empty array (new String[0]) and the toArray method will create an array of the right size. If you use set.toArray(new String[set.size()]);, the toArray method uses your array and returns it, so only one array is created. – assylias Jun 19 '13 at 22:00

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