如何在Scala中從a => b => c獲取（a，b）=> c？

Question

如果我有：

val f : A => B => C 

這是簡寫：

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

我如何獲得一個功能g與簽名：

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

（IE）

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

根據f？是對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

擴展retonym的答案，爲了完整性還提供了功能對象

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) 

對於這兩項操作相反的功能，讓你可以倒着寫上面的，如果你想

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

爲了圓滿答案，雖然有一個庫方法可以做到這一點，但它也可能是有益的：

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> 

或一般，

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

Answer 4

類似於雷克斯克爾的答案，但更容易閱讀。

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(_)(_)