实现更高阶函数,在scala中执行currying

时间:2012-12-10 01:22:40

标签: scala currying higher-order-functions

我的一位同事给我发了一个问题如下:

  

实现执行currying的HOF(高阶函数)   你的功能签名如下:

def curry[A,B,C](f:(A,B) => C) : A => B => C
  

同样,实现一个执行uncurrying的函数,如下所示:

def uncurry[A,B,C](f:A => B => C): (A,B) => C

我理解currying的方式是,如果你有一个带有多个参数的函数,你可以重复将函数应用于每个参数,直到得到结果。

因此f:(A,B) => C的某些内容变为A => f(A,_) => f(B) ????

并且不应该将此应用程序合并到一个函数中,如下所示:

f:A=>B=>C将是f(A,B)

也许我只是对这里的语法感到困惑,但是如果有人能指出我在这里缺少的东西会很棒。

由于

2 个答案:

答案 0 :(得分:12)

希望这个带有大量注释的完整工作示例很容易理解。如果您有任何疑问,请回复。

您可以通过将其放入Scala解释器来执行此代码。

// Here's a trait encapsulating the definition your coworker sent.
trait Given {
  def curry[A,B,C](f:(A,B) => C) : A => B => C
  def uncurry[A,B,C](f:A => B => C): (A,B) => C
}

object Impl extends Given {
  // I'm going to implement uncurry first because it's the easier of the
  // two to understand.  The bit in curly braces after the equal sign is a
  // function literal which takes two arguments and applies the to (i.e.
  // uses it as the arguments for) a function which returns a function.
  // It then passes the second argument to the returned function.
  // Finally it returns the value of the second function.
  def uncurry[A,B,C](f:A => B => C): (A,B) => C = { (a: A, b: B) => f(a)(b) }

  // The bit in curly braces after the equal sign is a function literal
  // which takes one argument and returns a new function.  I.e., curry()
  // returns a function which when called returns another function
  def curry[A,B,C](f:(A,B) => C) : A => B => C = { (a: A) => { (b: B) => f(a,b) } }
}

def add(a: Int, b: Long): Double = a.toDouble + b
val spicyAdd = Impl.curry(add)
println(spicyAdd(1)(2L)) // prints "3.0"
val increment = spicyAdd(1) // increment holds a function which takes a long and adds 1 to it.
println(increment(1L)) // prints "2.0"
val unspicedAdd = Impl.uncurry(spicyAdd)
println(unspicedAdd(4, 5L)) // prints "9.0"

一个较少数字的例子怎么样?

def log(level: String, message: String) { 
  println("%s: %s".format(level, message)) 
} 
val spicyLog = Impl.curry(log) // spicyLog's type is String => Unit
val logDebug = spicyLog("debug") // This new function will always prefix the log
                                 // message with "debug".
val logWarn = spicyLog("warn") // This new function will always prefix the log 
                               // message with "warn".
logDebug("Hi, sc_ray!") // prints "debug: Hi, sc_ray!"
logWarn("Something is wrong.") // prints "warn: Something is wrong."

<强>更新 您回答问“编译​​器如何评估a => b => f(a,b)等表达式”。嗯,它没有。至少在同事的代码片段中定义的东西是无法编译的。但是,一般情况下,如果您看到A => B => C形式的某些内容,则表示“将A作为参数的函数;它返回一个函数,该函数将B作为参数并返回C。”

答案 1 :(得分:8)

我不确定我真的理解你的问题 - 你想知道什么,除了实际的实施?如上所述,它应该是微不足道的:

def curry[A,B,C](f:(A,B) => C): A => B => C = 
  a => b => f(a,b)

a => b => f(a,b)的含义是“一个参数a的函数,其返回值为b => f(a,b),这又是一个参数b,其返回值是您执行的f(a,b)(其类型为C)“

a => b => f(a, b)如果有帮助可以写得更详细一点吗?

 { (a: A) => {           // a function of *one* argument, `a`
      (b: B) => {        // a function of *one* argument, `b`
         f(a, b)         // whose return value is what you get of you execute `f(a,b)` (whose type is `C`)
      }
   }
 }

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

(a, b) => f(a)(b)表示“两个参数(a, b)的函数,其返回值是您第一次将a应用于HoF时获得的值{ {1}},返回一个函数,该函数反过来使用f来返回b“。

这有帮助吗?