如果我有:
val f : A => B => C
这是以下的简写:
val f : Function1[A, Function1[B, C]]
如何获取带有签名的函数g
:
val g : (A, B) => C = error("todo")
(即)
val g : Function2[A, B, C] //or possibly
val g : Function1[(A, B), C]
在f
方面
发布于 2010-08-11 17:20:12
scala> val f : Int => Int => Int = a => b => a + b
f: (Int) => (Int) => Int = <function1>
scala> Function.uncurried(f)
res0: (Int, Int) => Int = <function2>
发布于 2010-08-11 18:29:36
为了完备性,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对象上也提供了这两种操作的反向函数,因此如果您愿意,可以向后编写上面的代码
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
发布于 2010-08-11 22:01:15
为了完善答案,尽管有一个库方法可以做到这一点,但手动完成也可能是有指导意义的:
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>
或者一般情况下,
def uncurry[A,B,C](f: A=>B=>C): (A,B)=>C = {
(a: A, b: B) => f(a)(b)
}
https://stackoverflow.com/questions/3456864
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