Scala, Pattern matching with no extraction

Question

Im looking to create a pattern matcher which doesnt require value extraction, and havent found a very satisfying way of doing this. Suppose I have the following illustrative example:

I want to match on Any being an Int

a:Any match {
  case IsAnInt => println(s"$a is an int")
}

Now the closest I can get is either boolean

a:Any match {
  case IsAnInt(true) => println(s"$a is an int")
}

object IsAnInt { 
  def unapply(a:Any):Option[Boolean] = Some(a.isInstanceOf[Int]) 
}

or unit

a:Any match {
  case IsAnInt(()) => println(s"$a is an int")
}

object IsAnInt { 
  def unapply(a:Any):Option[Unit] = if (a.isInstanceOf[Int]) Some(()) else None
}

Which I can do, but isnt nearly as cool... Is there any trick Im not aware of to achieve the first case?

(just to clarify the example is a simplification, I'm not looking for ways to identify an int)

elm · Accepted Answer · 2016-05-12 11:49:03Z

3

Try pattern matching on the type like this,

(1: Any) match {
  case v: Int => println(s"$v is an int")
  case ______ => println("not an int")
}

answered May 12, 2016 at 11:49

elm

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Comments

alextsc · Accepted Answer · 2016-05-12 11:57:28Z

2

You can use a boolean extractor, i.e. an extractor that just returns a boolean (instead of an Option) to indicate whether it matches:

object IsAnInt {
  def unapply(a : Any) : Boolean = a.isInstanceOf[Int]
}

val x : Any = 3
x  match {
  case IsAnInt() => println("Is an int")
  case _ => println("Is not an int")
}

You still have the parenthesis to seperate matching the object from the extractor though.

answered May 12, 2016 at 11:57

alextsc

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