在Scala中定义从字符串到函数的映射

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问题描述

我正在尝试使用键定义一个Map文字： String ，value：（Any）=> String 。我尝试以下，但是得到一个语法错误：

I am trying to define a Map literal with key: String, value: (Any)=>String. I tried the following, but get a syntax error:

def foo(x: Int): String = /...
def bar(x: Boolean): String = /...
val m = Map[String, (Any) => String]("hello" -> foo, "goodbye" -> bar)

推荐答案

有趣的是，没有人实际上给出了将的类型。这是一个例子：

Funny that no one actually gave a type that would work. Here's one such:

def foo(x: Int): String = x.toString
def bar(x: Boolean): String = x.toString
val m = Map[String, (Nothing) => String]("hello" -> foo, "goodbye" -> bar)

为什么它以这种方式工作是因为 Function1 在输入上是对比变体，所以（Nothing）=> String 是（Int）=>的超类串。它也是输出上的变体，所以（Nothing）=>任何将是任何其他 Function1 的超类。

The reason why it works this way is because Function1 is contra-variant on the input, so (Nothing) => String is a superclass of (Int) => String. It is also co-variant on the output, so (Nothing) => Any would be a superclass to any other Function1.

当然可以不要这样用。没有清单，你甚至不能发现原始类型的 Function1 是什么。你可以尝试这样的东西：

Of course, you can't use it like that. Without manifests, you can't even uncover what the original type of Function1 is. You could try something like this, though:

def f[T : Manifest](v: T) = v -> manifest[T]
val m = Map[String, ((Nothing) => String, Manifest[_])]("hello" -> f(foo), "goodbye" -> f(bar))

val IntManifest = manifest[Int]
val BooleanManifest = manifest[Boolean]
val StringManifest = manifest[String]
m("hello")._2.typeArguments match {
    case List(IntManifest, StringManifest) =>
        m("hello")._1.asInstanceOf[(Int) => String](5)
    case List(BooleanManifest, StringManifest) =>
        m("hello")._1.asInstanceOf[(Boolean) => String](true)
    case _ => "Unknown function type"
}

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SuperClass