Scala 如何获得(a,b)=>;c从a=>;b=>;在斯卡拉?
如果我有:Scala 如何获得(a,b)=>;c从a=>;b=>;在斯卡拉?,scala,functional-programming,currying,Scala,Functional Programming,Currying,如果我有: val f : A => B => C 这是以下的简写: val f : Function1[A, Function1[B, C]] 如何获得带有签名的函数g: val g : (A, B) => C = error("todo") (即) 在f?scala>valf:Int=>Int=>Int=a=>b=>a+b方面 scala> val f : Int => Int => Int = a => b => a + b f: (I
val f : A => B => C
这是以下的简写:
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)
在f
?scala>valf:Int=>Int=>Int=a=>b=>a+b方面
scala> val f : Int => Int => Int = a => b => a + b
f: (Int) => (Int) => Int = <function1>
scala> Function.uncurried(f)
res0: (Int, Int) => Int = <function2>
f:(Int)=>(Int)=>Int=
scala>函数。未载波(f)
res0:(Int,Int)=>Int=
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)
函数对象上还提供了这两个操作的反向函数,因此,如果愿意,可以将上述内容向后写入
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
让我们来总结一下答案,虽然有一种库方法可以做到这一点,但手工操作也可能有一定的指导意义:
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>
与Rex Kerr的答案相似,但更易于阅读
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(_)(_)
奇怪的是,
function本身没有uncurried
方法?要在Function1上有uncurried方法,您需要将其可接受的目标限制为返回函数的函数。也就是说,函数类型为Function1[A,Function1[B,C]]。这可能可以通过广义类型约束实现,但这些约束直到Scala 2.8才可用。
def uncurry[A,B,C](f: A=>B=>C): (A,B)=>C = {
(a: A, b: B) => f(a)(b)
}
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(_)(_)