有没有更好的方法在Scala中提升PartialFunction？
我偶尔会遇到以下模式，我基本上有一个
PartialFunction[SomeType,AnotherType]
Function[SomeType,Option[AnotherType]
def f(s:SomeType):Option[AnotherType] = s match {
case s1:SubType1 => Some(AnotherType(s1.whatever))
case s2:SubType2 => Some(AnotherType(s2.whatever))
case _ => None
}
Some
def f(s:SomeType):Option[AnotherType] = pf.lift(s)
def pf:PartialFunction[SomeType,AnotherType] = {
case s1:SubType1 => AnotherType(s1.whatever)
case s2:SubType2 => AnotherType(s2.whatever)
}
def f:Function[SomeType,Option[AnotherType]] = {
case s1:SubType1 => AnotherType(s1.whatever)
case s2:SubType2 => AnotherType(s2.whatever)
}.lift
没有找到相关结果
已邀请:
2 个回复
抵浮细
在对象scala.PartialFunction中。来自scaladoc：def onlyInt(v: Any): Option[Int] = condOpt(v) { case x: Int => x }辽躺
对象的方法，当你也可以直接创建它时：val pf: PartialFunction[SomeType,AnotherType] = { case s1:SubType1 => AnotherType(s1.whatever) case s2:SubType2 => AnotherType(s2.whatever) }val pf = { case s1:SubType1 => AnotherType(s1.whatever) case s2:SubType2 => AnotherType(s2.whatever) } : PartialFunction[SomeType,AnotherType]的确切签名。我知道感觉应该可以推断出来，但是，唉，可悲的是并非如此！ 使用归属版本，您可以在同一个地方定义和提升所有内容：val pf = ({ case s1:SubType1 => AnotherType(s1.whatever) case s2:SubType2 => AnotherType(s2.whatever) } : PartialFunction[SomeType,AnotherType]).lift是更好的解决方案，因为它允许推理器为你完成大部分工作，留下更清晰的代码:)