如何从Scala中的=> b => c得到(a,b)=> c？

Question

如果我有:

val f : A => B => C

这是以下的简写:

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

如何获得g带签名的功能:

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

(即)

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

在f？

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,扩展了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)

Function对象也提供了这两个操作的相反函数,因此如果你愿意的话你可以向后写这些函数

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,虽然有一种库方法可以做到这一点,但手动完成也可能是有益的:

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)
}