How to convert X => Option[R] to PartialFunction[X,R]

Question

As long as we have a PartialFunction[X,R] it\'s very easy to convert it to a function returning Option[R], e.g.

def pfToOptf[X, R](         


        

Answer 1

How about this:

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scala> def optfToPf[X,R](f: X => Option[R]): PartialFunction[X,R] = x => f(x) match {
     |     case Some(r) => r
     | }
optfToPf: [X,R](f: (X) => Option[R])PartialFunction[X,R]

scala>

Answer 2

Starting Scala 2.9, Function.unlift does precisely this:

def unlift[T, R](f: (T) => Option[R]): PartialFunction[T, R]

Turns a function T => Option[R] into a PartialFunction[T, R].

Answer 3

I suppose you could override apply and isDefinedAt by hand, but I'd do it the way you find ugly.

def optfToPf[X,R](f: X => Option[R]) = new PartialFunction[X,R] {
  def apply(x: X): R = f(x).get
  def isDefinedAt(x: X): Boolean = f(x) != None
}

Testing:

scala> val map = Map(1 -> 2)
map: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2)

scala> map(1)
res0: Int = 2

scala> def mapOpt(key: Int) = map.get(key)
mapOpt: (key: Int)Option[Int]

scala> mapOpt(1)
res1: Option[Int] = Some(2)

scala> mapOpt(2)
res2: Option[Int] = None

scala> val mapPf = optfToPf(mapOpt _)
mapPf: java.lang.Object with PartialFunction[Int,Int] = <function1>

scala> mapPf.isDefinedAt(2)
res3: Boolean = false

scala> mapPf.isDefinedAt(1)
res4: Boolean = true

scala> mapPf(1)
res5: Int = 2

scala> mapPf(2)
java.util.NoSuchElementException: None.get