Reaction to Cedric Beust’s equals challenge

February 17, 2013 4 minute read

This week, I got nerd sniped by Cedric Beust’s latest coding challenge:

A School has either a name (“Stanford”) or a nickname (“UCSD”) or both.

Your task is to write equals() and hashCode() methods for the School class that satisfy the following requirement: two Schools are identical if they either have the same name or the same nickname.

I wrote EqualsVerifier; I should be able to do this! I came up with the following, which I thought was pretty ironclad:

public final class School {
    private final String name;
    private final String nickname;

    public School(String name, String nickname) {
        this.name = name;
        this.nickname = nickname;
    }

    @Override
    public boolean equals(Object obj) {
        if (!(obj instanceof School)) {
            return false;
        }
        School other = (School)obj;
        return nullSafeEqual(name, other.name) ||
                nullSafeEqual(nickname, other.nickname);
    }

    private boolean nullSafeEqual(String a, String b) {
        return a == null ? b == null : a.equals(b);
    }
    
    @Override
    public int hashCode() {
        return 42;
    }

    @Override
    public String toString() {
        return "School: name=" + name + ", nickname=" + nickname;
    }
}

I even added a toString()! Obviously, I wasn’t really happy about the hashCode() method, but I wasn’t really sure how to write it. Essentially, this implementation will turn your efficient hash collection with O(1) lookup into a list with O(n) lookup. So yeah, that’s pretty bad. But at least it meets the contract, so I thought I’d fix that later; first I wanted to see if my equals() was correct. So I defined some tests with EqualsVerifier:

@Test
public void testEquals() {
    School one = new School("A", "1");
    School two = new School("A", "2");
    School three = new School("B", "2");

    School x = new School("X", "0");

    EqualsVerifier.forRelaxedEqualExamples(one, two)
            .andUnequalExample(x)
            .verify();

    EqualsVerifier.forRelaxedEqualExamples(two, three)
            .andUnequalExample(x)
            .verify();
}

So far so good! I had to use the slightly verbose forRelaxedEqualExamples() mode of EqualsVerifier here, because the regular case is meant for classes that are equal only when all their fields are equal, which is obviously not the case here: the equality relation defined by Beust is more relaxed – hence the name.

Then I decided to try and combine the two calls to EqualsVerifier:

EqualsVerifier.forRelaxedEqualExamples(one, two, three)
        .andUnequalExample(x)
        .verify();

And that’s where my test failed: Precondition: not all objects are equal. And that’s true: one and three aren’t equal to each other. But they should be, because the contract clearly states:

It is transitive: for any non-null reference values x, y, and z, if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) should return true.

So there you have it: this equality relation isn’t transitive. Therefore, it’s not possible to write an equals() method that follows Beust’s requirement, and still meet the contract. Was the challenge a trick question?

This also explains why I had trouble with my hashCode() method: equal objects must always return the same hashCode. So if one and three are somehow equal, without having anything in common except the link with two, then the only possible hashCode is a constant.

Now you might ask: so what if my equals() method isn’t transitive? If this is the kind of equality semantics that I want, then why can’t I have it? It’s a free country!

And that’s true: you’re free to break the equals contract. But, as with any contract, if you break it, there will be consequences. In this case, the consequences will be subtle, hard-to-track bugs. One of the commenters had a good example of one such bug. Consider the following code:

Set<School> red = new HashSet<School>();
red.add(one);
red.add(two);
red.add(three);

Set<School> black = new HashSet<School>();
black.add(two);
black.add(one);
black.add(three);

System.out.println("Collection red has " + red.size() + " elements.");
System.out.println("Collection black has " + black.size() + " elements.");

Can you guess what that prints? Actually, you can’t, because it depends on the implementation of HashSet. When you add a key to a HashSet that’s already present in the set, the implementation can do two things. Either, it keeps the original one, in which case it will print:

Collection one has 2 elements.
Collection two has 1 elements.

Or it replaces the old one with the new one, in which case the order will be reversed.

The moral of the story: writing a good equals method is hard; don’t try to make it fancy.

If you want to have a non-transitive equality test, that’s fine, but don’t abuse equals() to achieve that. There are other ways. One of them, as Beust points out in his follow-up post, is to add an id field to the School class and use that to base equals() and hashCode() on. Another way could be to simply add another method to the class:

public boolean isSameSchool(School other) {
    return nullSafeEqual(name, other.name) ||
            nullSafeEqual(nickname, other.nickname);
}

It even looks a lot better, because you don’t need all that type checking stuff anymore.

P.S.

Just a few paragraphs ago I was saying how I needed to use EqualsVerifier’s forRelaxedEqualExamples mode, because of the relaxed nature of this equality relation, and that the regular mode obviously wouldn’t work. I don’t know what made me decide to test that, anyway:

@Test
public void obviouslyThisTestShouldNotPass() {   // obviously!
    EqualsVerifier.forClass(School.class).verify();
}

It passed.

I’m going to have to fix that…

Jan Ouwens

P.S.