Java: Only check hashCode in equals() of immutable object

Question

I have an immutable object, for example a node in the Cartesian space. The class is immutable, so I cache the hashCode for very fast hashing.

private final int hashCode;

private final double x, y, z;

public Node(final double x, final double y, final double z)
{
    this.x = x;
    this.y = y;
    this.z = z;
    this.hashCode = Objects.hashCode(this.x, this.y, this.z);
}

@Override
public boolean equals(final Object obj)
{
    if (this == obj) { return true; }
    if (obj == null) { return false; }
    if (!(obj instanceof Node)) { return false; }
    final Node other = (Node) obj;
    return Objects.equal(this.x, other.x) && Objects.equal(this.y, other.y) && Objects.equal(this.z, other.z);
}

@Override
public int hashCode()
{
    return this.hashCode;
}

Since the hashCode is unique and dependent on all fields of the class AND the class is Immutable, would it be correct to only check Node equality based on the hashCode?

@Override
public boolean equals(final Object obj)
{
    if (this == obj) { return true; }
    if (obj == null) { return false; }
    if (!(obj instanceof Node)) { return false; }
    final Node other = (Node) obj;
    return this.hashCode == other.hashCode;
}

This passes all Unit Tests I have written about the properties of equals() and hashCode() and their interaction, but perhaps there is something I am missing?

Note: Objects.hashCode() and Objects.equal() are Guava classes helpful for the respective methods.

For reference, as also mentioned here: stackoverflow.com/a/7417858/713659 , you can check for hashCode first before going into more expensive equality checks, but you cannot skip other checks. — Alexander Karatarakis
– Alexander Karatarakis, Commented Apr 18, 2012 at 0:36
FYI, there's no need to use Objects.equal in the equals implementation, since you're comparing primitives. == will be faster and simpler. — Louis Wasserman
– Louis Wasserman, Commented Apr 18, 2012 at 0:50

SLaks · Accepted Answer · 2012-04-18 00:24:40Z

17

Nope; that won't work.

You have 2³² possible hashcodes and 2¹⁹² possible values.

answered Apr 18, 2012 at 0:24

SLaks

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Comments

bertvh · Accepted Answer · 2012-09-05 18:10:02Z

2

No, but..

I guess you could check the hashcode to see whether objects are not equal and gain some performance there:

public boolean equals(final Object obj) {
   if (this == obj) { return true; }
   if (!(obj instanceof Node)) { return false; }
   final Node other = (Node) obj;

   if (this.hashCode != other.hashCode) {
      return false; // If hashcodes differ, we're sure the objects are not equal
   }
   // remainder of the actual equals implementation
}

Of course this will only improve performance in case most of your comparisons yield false. In case of equal objects, this will bring a performance penalty. In your example (comparing just three values), I wouldn't recommend this.

answered Sep 5, 2012 at 18:10

bertvh

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