如何获取（A，B）=> C来自A => B =>在斯卡拉C吗？

Question

如果我有：

val f : A => B => C

这是简写：

val f : Function1[A, Function1[B, C]]

我如何获得一个功能g与签名：

val g : (A, B) => C = error("todo")

（即）

val g : Function2[A, B, C] //or possibly
val g : Function1[(A, B), C]

在f方面？

Answer 1

scala> val f : Int => Int => Int = a => b => a + b
f: (Int) => (Int) => Int = <function1>

scala> Function.uncurried(f)
res0: (Int, Int) => Int = <function2>

Answer 2

扩展retonym的回答，为了完整性

val f : Int => Int => Int = a => b => a + b
val g: (Int, Int) => Int = Function.uncurried(f)
val h: ((Int, Int)) => Int = Function.tupled(g)

还提供了功能对象上用于这两种操作的相反的功能，所以可以编写上述向后，如果你希望

val h: ((Int, Int)) => Int =  x =>(x._1 + x._2)
val g: (Int, Int) => Int = Function.untupled(h)
val f : Int => Int => Int = g.curried  //Function.curried(g) would also work, but is deprecated. Wierd

Answer 3

只是为了完善了答案，虽然有一个库的方法来做到这一点，它也可能是有益的手工做到这一点：

scala> val f = (i: Int) => ((s: String) => i*s.length)
f: (Int) => (String) => Int = <function1>

scala> val g = (i: Int, s: String) => f(i)(s)
g: (Int, String) => Int = <function2>

或者在一般情况下，

def uncurry[A,B,C](f: A=>B=>C): (A,B)=>C = {
  (a: A, b: B) => f(a)(b)
}

Answer 4

类似于雷克斯克尔的答案，但更容易阅读。

type A = String
type B = Int
type C = Boolean

val f: A => B => C = s => i => s.toInt+i > 10

val f1: (A, B) => C = f(_)(_)