inglesp
inglesp

Reputation: 3357

Java: Converting a set to an array for String representation

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?

Upvotes: 32

Views: 103955

Answers (6)

coobird
coobird

Reputation: 161032

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.

Upvotes: 33

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.

Upvotes: 0

David
David

Reputation: 159

As dfa mentioned, you can just replace:

System.out.println(array);

with...

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

Upvotes: 2

dfa
dfa

Reputation: 116442

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[][]).

Upvotes: 10

Reginaldo
Reginaldo

Reputation: 907

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

Upvotes: 5

jpalecek
jpalecek

Reputation: 47770

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.

Upvotes: 1

Related Questions