equals() & hashCode()

Default is identity

Object's default equals() is just == (reference identity), and default hashCode() is derived from that identity. If two objects should be "equal" by their content (e.g. two Point(1,2)), you must override both methods yourself.

The contract

The golden rule: if a.equals(b) is true, then a.hashCode() == b.hashCode() must also be true. Break this and objects silently vanish inside HashMap/HashSet lookups. Use the same fields in both methods.

public final class Point {
    private final int x, y;
    // ... constructor, getters ...

    @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);
    }
}

Quick check

Quick check: You override equals() but forget hashCode(). What breaks?

• Nothing, they're independent
• Compilation fails
• "Equal" objects can land in different HashMap buckets and lookups fail
• equals() itself throws at runtime