The Object Class: equals, hashCode, toString

Every Java class extends java.lang.Object, and inherits its methods. Three of them you will override in almost every data class: toString(), equals() and hashCode(). Getting the last two wrong causes some of the most confusing bugs in Java – objects that "disappear" from a HashSet or HashMap.

What Object gives you

The problem with the defaults

import java.util.HashSet;
import java.util.Set;

public class Main {
    public static void main(String[] args) {
        Point a = new Point(2, 3);
        Point b = new Point(2, 3);
        System.out.println(a.equals(b));          // false: different objects
        Set<Point> set = new HashSet<>();
        set.add(a);
        set.add(b);
        System.out.println(set.size());           // 2 "duplicates"
        System.out.println(a.toString().startsWith("Point@"));
    }
}

class Point {
    final int x, y;
    Point(int x, int y) { this.x = x; this.y = y; }
}

false
2
true

Two points with the same coordinates should be equal. For that we override all three methods.

A correct implementation

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Objects;
import java.util.Set;

public class Main {
    public static void main(String[] args) {
        Point a = new Point(2, 3), b = new Point(2, 3);
        System.out.println(a.equals(b) + " " + (a.hashCode() == b.hashCode()));
        Set<Point> set = new HashSet<>(Set.of(a));
        System.out.println(set.contains(b) + " " + set.size());
        Map<Point, String> labels = new HashMap<>();
        labels.put(a, "start");
        System.out.println(labels.get(new Point(2, 3)));
        System.out.println(a);
    }
}

class Point {
    private final int x, y;

    Point(int x, int y) { this.x = x; this.y = y; }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;                         // same object
        if (!(o instanceof Point other)) return false;      // null or another type
        return x == other.x && y == other.y;                // compare the fields that matter
    }

    @Override
    public int hashCode() {
        return Objects.hash(x, y);                          // same fields as equals
    }

    @Override
    public String toString() {
        return "Point(" + x + ", " + y + ")";
    }
}

true true
true 1
start
Point(2, 3)

The equals contract

The Java documentation requires equals to be:

• Reflexive: a.equals(a) is true.
• Symmetric: if a.equals(b) then b.equals(a).
• Transitive: if a equals b and b equals c, then a equals c.
• Consistent: repeated calls give the same answer while the objects don't change.
• a.equals(null) is always false.

Why hashCode must match equals

Hash-based collections first use hashCode() to pick a bucket, then use equals() to compare only inside that bucket.

If equal objects produced different hash codes, Java would look in the wrong bucket and never find the match:

import java.util.HashSet;
import java.util.Set;

public class Main {
    public static void main(String[] args) {
        Set<Badge> set = new HashSet<>();
        set.add(new Badge("JAVA"));
        System.out.println(new Badge("JAVA").equals(new Badge("JAVA")));
        System.out.println(set.contains(new Badge("JAVA")));   // almost always false!
    }
}

class Badge {
    final String code;
    Badge(String code) { this.code = code; }

    @Override
    public boolean equals(Object o) {
        return o instanceof Badge b && code.equals(b.code);
    }
    // hashCode() NOT overridden – breaks hash collections
}

true
false

Helpful utilities

import java.util.Objects;

public class Main {
    public static void main(String[] args) {
        String a = null, b = "java";
        System.out.println(Objects.equals(a, b));        // null-safe
        System.out.println(Objects.equals(null, null));
        System.out.println(Objects.hash("Asha", 21) == Objects.hash("Asha", 21));
        System.out.println(Objects.requireNonNullElse(a, "default"));
    }
}

false
true
true
default

Practice

Summary

Interview practice: Java OOP in depth.