如何定义匿名通用 Scala 函数? [英] How can I define an anonymous generic Scala function?

问题描述

假设我有这个:

val myAnon:(Option[String],String)=>String = (a:Option[String],defVal:String) => {
  a.getOrElse(defVal)
}

不要介意函数的作用.有没有办法让它通用，所以我可以有一个 Option[T]?

Don't mind what the function does. Is there anyway of making it generic, so I can have an Option[T]?

推荐答案

总结一下那个答案:不，你不能让 匿名 函数成为通用的，但是你可以明确地将你的函数定义为扩展 Function0、Function1、Function2 等特征之一并从这些特征定义应用函数的类.那么你定义的类可以是泛型的.这是原始文章的摘录，可在此处一个>:

To summarize from that answer: No, you can't make anonymous functions generic, but you can explicitly define your function as a class that extends one of the Function0, Function1, Function2, etc.. traits and define the apply function from those traits. Then the class you define can be generic. Here is the excerpt from the original article, available here:

scala> class myfunc[T] extends Function1[T,String] {
     |     def apply(x:T) = x.toString.substring(0,4)
     | }
defined class myfunc

scala> val f5 = new myfunc[String]
f5: myfunc[String] = <function>

scala> f5("abcdefg")
res13: java.lang.String = abcd

scala> val f6 = new myfunc[Int]
f6: myfunc[Int] = <function>

scala> f6(1234567)
res14: java.lang.String = 1234

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