scala uanpply without parameter

Question

I am a student who studies Scala in korea. I am learning about pattern matching and unapply methods. The thing I am confused about is that Emergency object has a parameter in unapply method. I can't know the reason when I don't put the parameter in the match block.

object Solution {

  def main(args: Array[String]) {

    val number1 = "010-123-1234"
    val number2 = "119"
    val number3 = "포도먹은 돼지"
    val numberList = List(number1, number2, number3)

    for (number <- numberList) {
      number match {
        case Emergency() => println("긴급전화다")
        case Normal(number) => println("일반 전화다" + number)
        case _ => println("판단할 수 없습니다.")
      }
    }
  }
}

object Emergency {
  def unapply(number: String): Boolean = {
    if (number.length == 3 && number.forall(_.isDigit)) true
    else false
  }
}

object Normal {
  def unapply(number: String): Option[Int] = {
    try {
      Some(number.replaceAll("-", "").toInt)
    } catch {
      case _: Throwable => None
    }
  }
}

Answer 1

Notice that return types of to unapply methods are different.

Normal.unapply returns an Option . When you do case Normal(foo) , unapply is called, and, if it returns Some(number) , the match is successful, and the number is assigned to local variable foo , and if it returns None , match fails.

Emergency.unapply returns a Boolean , so case Emergency() succeeds if it returns true , and fails otherwise, but there is no result to assign in case of success, thus, no "parameter".

Answer 2

The parameter in unapply is the object on which you are matching.

In this case the number String variable is passed to Emergency.unapply , Normal.unapply etc.

This link explains things nicely:

https://danielwestheide.com/blog/2012/11/21/the-neophytes-guide-to-scala-part-1-extractors.html