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.

Answer 1

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

Answer 2

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.

Answer 3

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