equals() vs hashCode() ⚖️

Override one, you must override the other.

These two methods define object identity. Override equals() without hashCode() and your objects mysteriously vanish from HashMap and HashSet.

The contract

• If a.equals(b) is true, then a.hashCode() == b.hashCode() must hold.
• Equal hash codes do not require equality (collisions are fine).
• Both must use the same fields.

Do it the easy way

A record generates correct equals, hashCode, and toString for you:

record Point(int x, int y) {}

For a regular class, derive both from the same fields:

final class Point {
    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;
        if (!(o instanceof Point p)) return false;
        return x == p.x && y == p.y;
    }
    @Override public int hashCode() {
        return Objects.hash(x, y);
    }
}

instanceof pattern matching (Java 16+) handles the null check and cast in one line.

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TL;DR: Always override equals() and hashCode() together using identical fields — or just use a record.

Reach for a record next time you need a value object and skip the boilerplate entirely.