Why it is implied that objects are equal if compareTo() returns 0?

Question

Let's have a class Person. Person has a name and height.

Equals and hashCode() takes into account only name. Person is comparable (or we implement comparator for it, does not matter which one). Persons are compared by height.

It seems reasonable to expect a situation where two different persons can have same height, but eg. TreeSet behaves like comapareTo()==0 means equals, not merely same size.

To avoid this, comparison can secondarily look at something else if size is the same, but then it cannot be used to detect same sized different objects.

Example:

import java.util.Comparator;
import java.util.HashSet;
import java.util.Objects;
import java.util.Set;
import java.util.TreeSet;

public class Person implements Comparable<Person> {

private final String name;
private int height;

public Person(String name,
        int height) {
    this.name = name;
    this.height = height;
}

public int getHeight() {
    return height;
}

public void setHeight(int height) {
    this.height = height;
}

public String getName() {
    return name;
}

@Override
public int compareTo(Person o) {
    return Integer.compare(height, o.height);
}

public boolean equals(Object obj) {
    if (obj == null) {
        return false;
    }
    if (getClass() != obj.getClass()) {
        return false;
    }
    final Person other = (Person) obj;
    if (!Objects.equals(this.name, other.name)) {
        return false;
    }
    return true;
}

public int hashCode() {
    int hash = 5;
    hash = 13 * hash + Objects.hashCode(this.name);
    return hash;
}

public String toString() {
    return "Person{" + name + ", height = " + height + '}';
}

public static class PComparator1 implements Comparator<Person> {

    @Override
    public int compare(Person o1,
            Person o2) {
        return o1.compareTo(o2);
    }
}

public static class PComparator2 implements Comparator<Person> {

    @Override
    public int compare(Person o1,
            Person o2) {
        int r = Integer.compare(o1.height, o2.height);
        return r == 0 ? o1.name.compareTo(o2.name) : r;
    }
}

public static void test(Set<Person> ps) {
    ps.add(new Person("Ann", 150));
    ps.add(new Person("Jane", 150));
    ps.add(new Person("John", 180));
    System.out.println(ps.getClass().getName());
    for (Person p : ps) {
        System.out.println(" " + p);
    }
}

public static void main(String[] args) {
    test(new HashSet<Person>());
    test(new TreeSet<Person>());
    test(new TreeSet<>(new PComparator1()));
    test(new TreeSet<>(new PComparator2()));
}
}

result:

java.util.HashSet
 Person{Ann, height = 150}
 Person{John, height = 180}
 Person{Jane, height = 150}

java.util.TreeSet
 Person{Ann, height = 150}
 Person{John, height = 180}

java.util.TreeSet
 Person{Ann, height = 150}
 Person{John, height = 180}

java.util.TreeSet
 Person{Ann, height = 150}
 Person{Jane, height = 150}
 Person{John, height = 180}

Do you have idea why it is so?

Answer 1

Extract from the java.util.SortedSet javadoc:

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.

Hence, in other words, SortedSet breaks (or "extends") the general contracts for Object.equals() and Comparable.compareTo. See the contract for compareTo:

It is strongly recommended, but not strictly required that (x.compareTo(y)==0) == (x.equals(y)). Generally speaking, any class that implements the Comparable interface and violates this condition should clearly indicate this fact. The recommended language is "Note: this class has a natural ordering that is inconsistent with equals."

Answer 2

It is recommended that compareTo only returns 0, if a call to equals on the same objects would return true:

The natural ordering for a class C is said to be consistent with equals if and only if e1.compareTo(e2) == 0 has the same boolean value as e1.equals(e2) for every e1 and e2 of class C. Note that null is not an instance of any class, and e.compareTo(null) should throw a NullPointerException even though e.equals(null) returns false.

(From the JDK 1.6 Javadocs)

Answer 3

TreeSet doesn't operate using hash codes and equality - it only operates on the basis of the comparator you give it. Note that the Javadoc states:

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.

In your case, your comparison *isn't consistent with equals, so your set doesn't obey the general contract of Set.

Why not just add more aspects to the comparison so that only equal elements compare with a result of 0?

Answer 4

you can fix it by using name for another comparison when the heights are equal

@Override
public int compareTo(Person o) {
    if(height == o.height)return name.compareTo(o.name);

    return Integer.compare(height, o.height);
}

since names are unique this will only return 0 if this.equals(o)

Answer 5

It is strongly recommended, but not strictly required that (x.compareTo(y)==0) == (x.equals(y)) [1]

So it's fine that you compare with a different criteria than the used on equals granted that you document it.

However it would be better if your compare by the same criteria and if needed provide a custom comparator which works on the new criteria ( height in the case of Person )

Answer 6

When you give Person a Comparator that compares instances on the height attribute of the Person, it really means that two Person instances are the same if they have the same height. You will have to make a Comparator that is specific for class Person.