Java: Converting a set to an array for String representation

Question

From Sun's Java Tutorial, I would have thought this code would convert a set into an array.

import java.util.*;

public class Blagh {
    public static void main(String[] args) {
        Set<String> set = new HashSet<String>();
        set.add("a");
        set.add("b");
        set.add("c");
        String[] array = set.toArray(new String[0]);
        System.out.println(set);
        System.out.println(array);
    }
}

However, this gives

[a, c, b]
[Ljava.lang.String;@9b49e6

What have I misunderstood?

Was the order of the elements accurate in this? I am having a similar situation where the order seems to have a mind of its own and not in the order that items were added. — electrichead
– electrichead, Commented Jan 24, 2012 at 3:45
Sets in general don't have a defined order, so you can't rely on the order you put the the elements into the set being the same as the order you get them out. Try the LinkedHashSet implementation of the Set interface if that's what you want; it adds this guarantee. — Frans
– Frans, Commented Jul 10, 2012 at 13:29

obataku · Accepted Answer · 2012-08-28 02:06:52Z

33

The code works fine.

Replace:

System.out.println(array);

With:

System.out.println(Arrays.toString(array));

Output:

[b, c, a]
[b, c, a]

The String representation of an array displays the a "textual representation" of the array, obtained by Object.toString -- which is the class name and the hash code of the array as a hexidecimal string.

edited Aug 28, 2012 at 2:06

obataku

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answered Jun 14, 2009 at 13:15

coobird

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4 Comments

coobird Over a year ago

Some parts of Java can be a little bit unfriendly.

dfa Over a year ago

he should replace System.out.println(array); not System.out.println(set);

obataku Over a year ago

@coobird I know this is over 3 years old, but it's best to use Arrays.toString over Arrays.asList followed by List.toString :-)

Matt Over a year ago

ps: note that when you call the "toArray" method, you shouldn't construct a zero length array. it's a waste, since it will internally have to create a new array to fit your data. If you do this: set.toArray(new String[set.size()]) you can save yourself an array allocation, as the "toArray" method will see that the passed array is large enough to hold the data. Directly from the javadocs: If the collection fits in the specified array, it is returned therein. Otherwise, a new array is allocated with the runtime type of the specified array and the size of this collection.

dfa · Accepted Answer · 2009-06-14 13:30:50Z

10

for the sake of completeness check also java.util.Arrays.toString and java.util.Arrays.deepToString.

The latter is particularly useful when dealing with nested arrays (like Object[][]).

answered Jun 14, 2009 at 13:30

dfa

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1 Comment

Quinn Taylor Over a year ago

I agree, it's good to not that these methods may be more efficient than creating an intermediate list via asList(), especially for large arrays. Note that deepToString() is most useful for multi-dimensional arrays and complex nesting.

Reginaldo · Accepted Answer · 2009-06-14 13:20:38Z

5

It's OK.

You are not seeing the array contents with System.out.println(array) because println calls object.toString() to get the bytes from an Object for output.

Since HashSet overrides the default toString() implementation, you can see the set contents with System.out.println(set);

As arrays do not override the default toString() (that gives the class name and some sort of identity hash code), you are getting the fuzzy [Ljava.lang.String;@9b49e6

Hope that helps

answered Jun 14, 2009 at 13:20

Reginaldo

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Comments

David · Accepted Answer · 2009-06-14 14:15:34Z

2

As dfa mentioned, you can just replace:

System.out.println(array);

with...

System.out.println(Arrays.toString(array));

answered Jun 14, 2009 at 14:15

David

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Comments

jpalecek · Accepted Answer · 2009-06-14 13:20:26Z

1

I don't think you have misunderstood anything; the code should work. The array, however, is not smart enough to print its contents in the toString method, so you'll have to print the contents with

for(String s : array) println(s);

or something like that.

answered Jun 14, 2009 at 13:20

jpalecek

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Comments

Thorbjørn Ravn Andersen · Accepted Answer · 2009-06-14 17:13:40Z

0

You have the correct result. Unfortunately the toString()-method on the array is still the original Object.toString() so the output is somewhat unusable per default but that goes for all arrays.

answered Jun 14, 2009 at 17:13

Thorbjørn Ravn Andersen

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