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
带有签名的函数:
val g : (A, B) => C = error("todo")
(IE)
val g : Function2[A, B, C] //or possibly
val g : Function1[(A, B), C]
方面f
?
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
带有签名的函数:
val g : (A, B) => C = error("todo")
(IE)
val g : Function2[A, B, C] //or possibly
val g : Function1[(A, B), C]
方面f
?
scala> val f : Int => Int => Int = a => b => a + b
f: (Int) => (Int) => Int = <function1>
scala> Function.uncurried(f)
res0: (Int, Int) => Int = <function2>
为了完整起见,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)
Function 对象上还提供了这两种操作的逆函数,因此如果您愿意,可以将上面的内容倒写
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
只是为了完善答案,虽然有一个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)
}
类似于 Rex Kerr 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(_)(_)