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

Asked 2009-12-15T15:41:52.717
24

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](f: PartialFunction[X,R])(x: X) =
    if (f.isDefinedAt(x)) Some(f(x))
    else None

However, what if the task is opposite: suppose I have a function f getting X as an argument and returning Option[R] as a result. And I want to make a PartialFunction[X,R] out of it. What is the best way?

What I've come up with looks pretty ugly to my taste:

def optfToPf[X,R](f: X => Option[R]) : PartialFunction[X,R] = {
    object extractor {
        def unapply(x: X): Option[R] = f(x)
    }

    { case extractor(r) => r }
}

Is there some better way I missed?

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1 Answer

12

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>
answered 2009-12-15T16:01:45.127

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