Java TreeSet with length Comparator bug?

Question

I have the below code which creates a TreeSet using a comparator based on string length.

public class TreeSetComparator {
    public static void main(String[] args) {
        SortedSet<String> sortedSet = new TreeSet<>(Comparator.comparing(String::length));
        sortedSet.addAll(Arrays.asList("aa", "bb", "aa"));
        System.out.println(sortedSet);
    }
}

To my surprise the output of the above is

[aa]

While I would expect

[aa, bb]

or

[bb, aa]

The "bb" part disappears, which seems to be contrary to the SortedSet contract. The comparator is supposed to only sort the elements and not determine their uniqueness, which is normally determined by equals.

On the other hand, if I enhance the comparator to always return non-zero for unequal items like below, only then do I get the correct results.

    SortedSet<String> sortedSet = new TreeSet<>(Comparator.comparing(String::length).reversed().thenComparing(String::toString));
    sortedSet.addAll(Arrays.asList("aa", "bb", "aa"));
    System.out.println(sortedSet);

The output now is [aa, bb] as I would expect.

Is the above a bug in the TreeSet implementation?

My environment is as follows:

mvn --version                                                                                                                                            21:40:22
Apache Maven 3.5.4 (1edded0938998edf8bf061f1ceb3cfdeccf443fe; 2018-06-17T19:33:14+01:00)
Maven home: /home/aaaa/.sdkman/candidates/maven/current
Java version: 10.0.2, vendor: Oracle Corporation, runtime: /usr/lib/jvm/java-10-jdk
Default locale: en_GB, platform encoding: UTF-8
OS name: "linux", version: "4.14.60-1-manjaro", arch: "amd64", family: "unix"

UPDATE

Here is a related post along with suggestions on how to fix the issue in a future version of Java: https://yesday.github.io/blog/2018/java-gotchas-sorted-set-ignores-the-equals-method.html

*The comparator is supposed to only sort the elements and not determine their uniqueness*... why do you think so? — shmosel, Aug 08 '18 at 20:46
It's a `Set`. Is guarantees uniqueness. Why do you think it would allow duplicates?! — Boris the Spider, Aug 08 '18 at 20:47
you probably want a comparator with a fallback comparison, like, if they have equal length, compare them lexicographically — Walter Tross, Aug 08 '18 at 20:49
@WalterTross OP figured this already and proposed it as solution later in his question. The question is just about **why** the first one *loses* one item. — Zabuzard, Aug 08 '18 at 20:50

Marvin · Accepted Answer · 2018-08-08T20:51:56.180

This it not a bug. At least not in TreeSet.

From the javadoc, emphasis by me:

Note that the ordering maintained by a set (whether or not an explicit comparator is provided) must be consistent with equals if it is to correctly implement the Set interface. (See Comparable or Comparator for a precise definition of consistent with equals.) This is so because the Set interface is defined in terms of the equals operation, but a TreeSet instance performs all element comparisons using its compareTo (or compare) method, so two elements that are deemed equal by this method are, from the standpoint of the set, equal. The behavior of a set is well-defined even if its ordering is inconsistent with equals; it just fails to obey the general contract of the Set interface.

So because "aa" and "bb" both have a length of 2 they are deemed equal by compareTo and thus by the TreeSet.

By definition, consistent with equals means:

The ordering imposed by a comparator c on a set of elements S is said to be consistent with equals if and only if c.compare(e1, e2)==0 has the same boolean value as e1.equals(e2) for every e1 and e2 in S.

score 0 · Answer 2 · answered Aug 08 '18 at 20:49

Looks like they're assuming that the comparator uses the same definition of equals as the equals method. From the SortedSet API:

Note that the ordering maintained by a sorted set (whether or not an explicit comparator is provided) must be consistent with equals if the sorted set is to correctly implement the Set interface. (See the Comparable interface or Comparator interface for a precise definition of consistent with equals.) This is so because the Set interface is defined in terms of the equals operation, but a sorted set performs all element comparisons using its compareTo (or compare) method, so two elements that are deemed equal by this method are, from the standpoint of the sorted set, equal. The behavior of a sorted set is well-defined even if its ordering is inconsistent with equals; it just fails to obey the general contract of the Set interface.

Java TreeSet with length Comparator bug?

2 Answers2

Linked

Related