Scala: Ignore case class field for equals/hascode?

Question

Is it possible to ignore a field of a case class in the equals/haschode method of the case class?

My use case is that I have a field that is essentially metadata for rest of the data in the class.

If you overwrite hashcode/equals? – user unknown Apr 29 '12 at 16:20 — user unknown, Apr 29 '12 at 16:20

score 95 · Accepted Answer · answered Apr 29 '12 at 16:48

95

Only parameters in the first parameter section are considered for equality and hashing.

scala> case class Foo(a: Int)(b: Int)
defined class Foo

scala> Foo(0)(0) == Foo(0)(1)
res0: Boolean = true

scala> Seq(0, 1).map(Foo(0)(_).hashCode)
res1: Seq[Int] = List(-1669410282, -1669410282)

UPDATE

To expose b as a field:

scala> case class Foo(a: Int)(val b: Int)
defined class Foo

scala> Foo(0)(1).b
res3: Int = 1

answered Apr 29 '12 at 16:48

retronym

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But `Foo(0)(0).b` doesn't work. It gives the error `value b is not a member of Foo`. – Ken Bloom Apr 29 '12 at 16:49
2

@KenBloom if you add "val" before "b" it will work: case class Foo(a: Int)(val b: Int) – czajah Jul 01 '14 at 16:55

score 7 · Answer 2 · answered Apr 29 '12 at 16:45

scala> :paste
// Entering paste mode (ctrl-D to finish)

case class Foo private(x: Int, y: Int) {
  def fieldToIgnore: Int = 0
}

object Foo {
  def apply(x: Int, y: Int, f: Int): Foo = new Foo(x, y) {
    override lazy val fieldToIgnore: Int = f
  }
}

// Exiting paste mode, now interpreting.

defined class Foo
defined module Foo

scala> val f1 = Foo(2, 3, 11)
f1: Foo = Foo(2,3)

scala> val f2 = Foo(2, 3, 5)
f2: Foo = Foo(2,3)

scala> f1 == f2
res45: Boolean = true

scala> f1.## == f2.##
res46: Boolean = true

You may override .toString if necessary.

score 2 · Answer 3 · edited Apr 29 '12 at 17:51

2

You can override the equals and hasCode methods in a case class

scala> :paste
// Entering paste mode (ctrl-D to finish)

case class Person( val name:String, val addr:String) {
  override def equals( arg:Any) = arg match {
    case Person(s, _) => s == name
    case _ => false
  }
  override def hashCode() = name.hashCode
}

// Exiting paste mode, now interpreting.

scala> Person("Andy", "") == Person("Andy", "XXX")
res2: Boolean = true

scala> Person("Andy", "") == Person("Bob", "XXX")
res3: Boolean = false

edited Apr 29 '12 at 17:51

Luigi Plinge

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answered Apr 29 '12 at 17:00

andy

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Not a very attractive option as [`.equals` is hard to get right](http://www.artima.com/lejava/articles/equality.html), and `.equals` and `.hashCode` together make for a lot of boilerplate. – missingfaktor Apr 29 '12 at 17:07
agree that equal/hashCode overrighting is error prone (as in java), just looks at sample code for an example. However all other methods, apply/unapply/toString/etc will be auto generated and will work as before. – andy Apr 29 '12 at 17:48
I'd say it's easy to get wrong, rather than hard to get right, if that makes any sense :) Have updated to something more correct – Luigi Plinge Apr 29 '12 at 17:52

score 0 · Answer 4 · answered Apr 23 '13 at 13:45

If you override toString in the base class it will not be overridden by the derived case classes. Here is an example:

sealed abstract class C {
  val x: Int
  override def equals(other: Any) = true
}

case class X(override val x: Int) extends C

case class Y(override val x: Int, y: Int) extends C

Than we you test:

scala> X(3) == X(4)
res2: Boolean = true

scala> X(3) == X(3)
res3: Boolean = true

scala> X(3) == Y(2,5)
res4: Boolean = true

Scala: Ignore case class field for equals/hascode?

4 Answers4

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