Scala choice in implementation options

Question

How to do it in Scala:

  sealed trait Option[+A] {
    def get: A
    def isEmpty: Boolean
    def map[B](f: A => B): Option[B] =
      if (isEmpty) None else Some(f(this.get))
  }
  object None extends Option[Nothing] {
    def isEmpty = true
    def get = throw new NoSuchElementException("None.get")
  }
  case class Some[+A](x: A) extends Option[A] {
    def isEmpty = false
    def get = x
  }

As I guess it is in the OOP world:

  sealed trait Option[+A] {
    def map[B](f: A => B): Option[B]
  }
  object None extends Option[Nothing] {
    def map[B](f: Nothing => B): Option[B] = this
  }
  case class Some[+A](get: A) extends Option[A] {
    def map[B](f: A => B): Option[B] = Some(f(get))
  }

What happened to the last one?

Functional programming in Scala uses matchin the Option[A]tag what is the third way (looks like Haskell, but why?) Why not use the subtype polytype?

UPDATE . The third way:

sealed trait Option[+A] {
  def map[B](f: A => B): Option[B] = this match {
    case None => None
    case Some(a) => Some(f(a))
  }
}
object None extends Option[Nothing] {
}
case class Some[+A](get: A) extends Option[A] {
}

Answer 1

, , isEmpty get, , Option, Some. , map , , map , , map .

Answer 2

, , scala , java-, , .
java-

val opt: Option[String] = ???
if (!opt.isEmpty) {
  //do something with opt.get
} else {
  //do some default...
}

, ( getOrElse).

@Victor Moroz this match { None => default_val; Some(v) => v }
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