Scala中的空部分函数

Question

在我看来{ case ... => ... },部分函数的语法至少需要一个case:

scala> val pf: PartialFunction[String, String] = { case "a" => "b" } 
pf: PartialFunction[String,String] = <function1>

scala> val pf: PartialFunction[String, String] = { }                
<console>:5: error: type mismatch;
 found   : Unit
 required: PartialFunction[String,String]
       val pf: PartialFunction[String, String] = { }
                                                 ^

那么,定义"空"部分函数的最佳方法是什么？难道还有比"手动"覆盖更好的办法isDefinedAt和apply？

Answer 1

Map是PartialFunction,所以你可以这样做:

val undefined: PartialFunction[Any, Nothing] = Map.empty

Answer 2

从Scala 2.10开始,您可以使用:

val emptyPf = PartialFunction.empty[String, String]

Answer 3

scala> def pfEmpty[A, B] = new PartialFunction[A, B] {
     |   def apply(a: A): B = sys.error("Not supported")
     |   def isDefinedAt(a: A) = false
     | }
pfEmpty: [A, B]=> java.lang.Object with PartialFunction[A,B]

scala> val f = pfEmpty[String, String]
f: java.lang.Object with PartialFunction[String,String] = <function1>

scala> f.lift
res26: (String) => Option[String] = <function1>

scala> res26("Hola")
res27: Option[String] = None

正如@didierd在评论中所说,由于参数差异,单个实例可以涵盖所有可能的参数类型.

scala> object Undefined extends PartialFunction[Any, Nothing] {
     |   def isDefinedAt(a: Any) = false
     |   def apply(a: Any): Nothing = sys.error("undefined")
     | }
defined module Undefined

scala> val f: PartialFunction[String, String] = Undefined
f: PartialFunction[String,String] = <function1>

scala> f.lift apply "Hola"
res29: Option[String] = None

Answer 4

从每个人那里窃取,可能的混合:

val undefined : PartialFunction[Any, Nothing] = {case _ if false =>
  sys.error("undefined")
}

Answer 5

我能想到的最短的一个:

{ case _ if false => "" }