我如何得到 (a, b) => c 来自 a => b =>斯卡拉中的c？

Question

如果我有：

val f : A => B => C

这是以下简写：

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

如何获得带有签名的函数g：（

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

即）

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

用f表示？

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

为了完整性，扩展 retonym 的答案

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)

这两个操作的逆函数也在 Function 对象上提供，因此如果您愿意，您可以向后编写上面的内容

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

只是为了完善答案，尽管有一个库方法可以执行此操作，但手动执行此操作也可能具有指导意义：

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>

或者一般来说，

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

Answer 4

与 Rex Kerr 的答案类似，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(_)(_)