How to convert X => Option[R] to PartialFunction[X,R]
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].
How about this:
Welcome to Scala version 2.8.0.r19650-b20091114020153 (Java HotSpot(TM) Client VM, Java 1.6.0_17).
Type in expressions to have them evaluated.
Type :help for more information.
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>
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