Java: Converting a set to an array

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?

Answer 1

The code works fine.

Replace:

System.out.println(array);

With:

System.out.println(Arrays.asList(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.

Answer 2

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.

Answer 3

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

Answer 4

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[][]).

Answer 5

As dfa mentioned, you can just replace:

System.out.println(array);

with...

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

Answer 6

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.