Scala 中的类型化函数和柯里化 [英] Typed Function and Currying in Scala

问题描述

在 Scala 中，假设我有一个这样的函数:

In Scala let's say I have a function like this:

def foo[R](x: String, y: () => R): R

所以我可以这样做:

val some: Int = foo("bar", { () => 13 })

有没有办法在不丢失"第二个参数的类型的情况下将其更改为使用函数柯里化?

Is there a way to change this to use function currying without "losing" the type of the second argument?

def foo[R](x: String)(y: () => R): R
val bar = foo("bar") <-- this is now of type (() => Nothing)
val some: Int = bar(() => 13) <-- doesn't work

推荐答案

senia 答案的变体，以避免结构化类型:

A variation on senia's answer, to avoid structural typing:

case class foo(x: String) extends AnyVal {
  def apply[R](y: () => R): R = y()
}

val bar = foo("bar")
val some: Int = bar(() => 13)
// Int = 13

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