Scala, partial functions

问题

Is there any way to create a PartialFunction except through the case statement?

I'm curious, because I'd like to express the following (scala pseudo ahead!)...

val bi = BigInt(_)
if (bi.isValidInt) bi.intValue

... as a partial function, and doing

val toInt : PartialFunction[String, Int] = {
    case s if BigInt(s).isValidInt => BigInt(s).intValue
}

seems redundant since I create a BigInt twice.

回答1:

Not sure I understand the question. But here's my attempt: Why not create an extractor?

object ValidBigInt {
  def unapply(s: String): Option[Int] = {
    val bi = BigInt(s)
    if (bi.isValidInt) Some(bi.intValue) else None
  }
}

val toInt: PartialFunction[String, Int] = {
  case ValidBigInt(i) => i
}

The other option is (and that may answer the question as to whether one can create PartialFunction other than with a case literal):

val toInt = new PartialFunction[String, Int] {
  def isDefinedAt(s: String) = BigInt(s).isValidInt
  def apply(s: String) = BigInt(s).intValue
}

However since the idea of a partial function is that it's only partially defined, in the end you will still do redundant things -- you need to create a big int to test whether it's valid, and then in the function application you create the big int again...

I saw a project at Github that tried to come around this by somewhat caching the results from isDefinedAt. If you go down to the benchmarks, you'll see that it turned out to be slower than the default Scala implementation :)

So if you want to get around the double nature of isDefinedAt versus apply, you should just go straight for a (full) function that provides an Option[Int] as result.

回答2:

I think you're looking for lift/unlift. lift takes a partial function and turns it into a function that returns an Option. Unlift takes a function with one argument that returns an option, and returns a partial function.

import scala.util.control.Exception._

scala> def fn(s: String) = catching(classOf[NumberFormatException]) opt {BigInt(s)}
fn: (s: String)Option[scala.math.BigInt]

scala> val fnPf = Function.unlift(fn)
fnPf: PartialFunction[String,scala.math.BigInt] = <function1>

scala> val fn = fnPf.lift
fn: String => Option[scala.math.BigInt] = <function1>

Closely related, you also want to look at this answer for information about cond and condOpt:

scala> import PartialFunction._
import PartialFunction._

scala> cond("abc") { case "def" => true }
res0: Boolean = false

scala> condOpt("abc") { case x if x.length == 3 => x + x }
res1: Option[java.lang.String] = Some(abcabc)

回答3:

You can write out a PartialFunction "longhand" if you'd like:

object pf extends PartialFunction[Int,String] {
  def isDefinedAt(in: Int) = in % 2 == 0

  def apply(in: Int) = {
    if (in % 2 == 0) 
      "even" 
    else 
      throw new MatchError(in + " is odd")
}

回答4:

Okay, I got this

import java.lang.NumberFormatException
import scala.util.control.Exception._

val toInt: PartialFunction[String, Int] = {
  catching(classOf[NumberFormatException]) opt BigInt(_) match {
    case Some(bi) if bi.isValidInt => bi.intValue
  }
}

回答5:

How about this?

val toInt: PartialFunction[String, Int] = (s: String) => BigInt(s) match {
  case bi if bi.isValidInt => bi.intValue
}

来源：https://stackoverflow.com/questions/5668053/scala-partial-functions

标签

scala

pattern-matching

partialfunction

extractor