Understanding Scala's Extractor - Pattern with Zero variable and Boolean Result

Question

Currently I am learning Scala Extractor and stuck in following confusion. I am unable to understand the following code. In the below pattern match, How UpperCase() return a String while the unapply method is designed to return Boolean?

  object UpperCase {
    def unapply(s: String): Boolean = s.toUpperCase == s
  }

  println(UpperCase.unapply("RAK")) //print boolean true or false.

  "RAK" match{
    case status @ UpperCase() => println("yes - "+ status) //How status holds RAK not boolean value?
    case _ => println("No")
  }

Answer 1

You are using a boolean extractor, that matches all values v for which x.unapply(v) yields true. The @ is a pattern binder, which binds the variable status to the value matched by the pattern.

In your example, the pattern match the String "RAK", which is bound to the variable status.