如何在Scala中从a => b => c获取（a，b）=> c？

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") 

（IE）

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

根据f？是对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

扩展retonym的答案，为了完整性还提供了功能对象

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) 

对于这两项操作相反的功能，让你可以倒着写上面的，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

为了圆满答案，虽然有一个库方法可以做到这一点，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

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(_)(_)