Unexpected output using comparator

Question

I have the following program

import java.util.*;
public class Test {
    public static void main(String[] args) {
        Integer[] array = { 3, 1, 4, 1, 5, 9 };
        Arrays.sort(array, new Comparator<Integer>() {
            public int compare(Integer i1, Integer i2) {
                return i1 < i2 ? -1 : (i2 > i1 ? 1 : 0);
            }
        });
        System.out.println(Arrays.toString(array));
    }
}

This gives me the output [3, 1, 4, 1, 5, 9]. Why?

asenovm · Accepted Answer · 2012-01-17 20:00:48Z

6

because i1 < i2 is the same as i2 > i1 - look what you've written in your compareTo method.

answered Jan 17, 2012 at 20:00

asenovm

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James Montagne · Accepted Answer · 2012-01-17 20:00:31Z

3

i1 < i2 
i2 > i1

These are the same.

answered Jan 17, 2012 at 20:00

James Montagne

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Comments

user949300 · Accepted Answer · 2012-01-17 20:11:05Z

1

You test for i1 < i2, and, if it fails, you test NOT for i1 > i2, but for i2 > i1.

Change your comparator to

   return i1 < i2 ? -1 : (i1 > i2) ? 1 : 0;

ADDED

In an attempt to contribute something new to the two other answers that said the same thing and beat me to the punch, most of the wrapper classes have built in compareTo() methods to save you from any thought. e.g., your comparator could just call Integer.compareTo(), i.e.

 return i1.compareTo(i2);

edited Jan 17, 2012 at 20:11

answered Jan 17, 2012 at 20:05

user949300

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1 Comment

G. Bach Over a year ago

Probably should be using compareTo() anyway since those magic numbers don't look healthy; +1 for that.

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