java comparing integers as strings with a comparator - odd results

Question

I'm attemping to compare some players with a comparator by the amount of runs they have obtained.

System.out.println("Comparing: " + p2.getRuns() + " and " + p1.getRuns());
int newRESULT = intConvert(p2.getRuns()).compareTo(intConvert(p1.getRuns()));
System.out.println("Returns: " + newRESULT);
return newRESULT;

However this returns:

Comparing: 25 and 0, Returns: 2

Comparing: 0 and 100, Returns: -1

Comparing: 25 and 100, Returns: 1

...and hence orders the players in the wrong order.

Should the first comparison not return 1, the second -1 and the last -1 as well?

intConvert:

     private static String intConvert(int x)
     {
      return "" + x;
     }

Note that `compareTo` doesn't have to return -1 or 1, it could return smaller or bigger values as well. Basically the convention is: < 0 means smaller, == 0 means equal and > 0 means greater. — Thomas, May 02 '12 at 17:37

score 3 · Accepted Answer · answered May 02 '12 at 17:40

I assume intConvert(...) converts an int to a String, and thus you get lexical comparisons which meahs "25" is greater than "100" because the first character is greater (2 > 1).

If you want to get correct comparisons stick to comparing ints or if you need to use a String create strings of equal length and fill in missings zeros at the front (e.g. 25 -> "025").

score 0 · Answer 2 · edited May 23 '17 at 11:59

To compare Numbers that are represented as String in a sensible way, you need to consistently format them all the same way.

Strings use lexical comparisons. This means "5" will be > "20" because 5 is > than 2. To get the logically expected output you need to format the numbers with some kind of formatter to make them lexically comparable.

"05" is < "20"

The simplest thing would be to use String.format() and a pattern to format all the numbers to Strings so they will compare consistently lexically.

String.format("%04d", yourInteger);

This will produce all ints that are passed in as left padded with 0 for 4 positions.

String.format("%04d", 5);

will produce 0005

Make the "%0Xd" where X is the number of digits you want it formatted to.

score 0 · Answer 3 · edited Jun 20 '20 at 09:12

You don't have to convert the numbers to strings just to sort, you can do something like:

class ComparePlayersByRuns implements Comparator<Player> {
    public int compareTo(Player p1, Player p2) {
        return p1.getRuns().compareTo(p2.getRuns());
    }
}

or in Java 8 and later all you need to create your comparator is:

Comparators.comparing(Player::getRuns);

And no, the compare isn't required to return 1, 0, or -1, the documentation says:

Returns:

a negative integer, zero, or a positive integer as the first argument is less than, equal to, or greater than the second.

java comparing integers as strings with a comparator - odd results

3 Answers3