可以在 Scala HList 上执行 Map [英] Can Map be performed on a Scala HList

问题描述

我现在已经完成了一些 HList 的实现.一个基于 Daniel Spiewak 的 High Wizardry in the Land of Scala 演讲，另一个基于 Apocalisp 博客中的一篇文章.目标是拥有一个异质列表，其中的主要类型不是异质的，而是更高的类型.例如:

I have done a few implementations of HList now. One based on Daniel Spiewak's High Wizardry in the Land of Scala talk and another based on a post in Apocalisp blog. The goal was to have a heterogenous list of which is not heterogenous in the primary type but rather the higher kind. For example:

val requests = Request[String] :: Request[Int] :: HNil

我将能够跨列表绘制地图以执行请求并生成更高类型的异构列表.所以:

I would be able to do a map across the list to perform the request and result in a heterogenous list of the higher kind. So:

requests.map(execute)

应该相等

String :: Int :: HNil

遗憾的是，我所有的尝试都得到了 Any 的 HList.这是最近尝试的代码:

Sadly all my attempts have resulted in an HList of Any. Here is the code from a recent attempt:

class Request[+Out](o:Out) {
  type O = Out

  def v:O = o
}

object HList {
  trait Func[-Elem,Out] {
    type Apply[E <: Elem] <: Out
    def apply[N <: Elem](e:N):Apply[N]
  }
  sealed trait HList[Base] {
    type Head <: Base
    type Tail <: HList[Base]
    type Map[Out,F <: Func[Base,Out]] <: HList[Out]
    def head:Head
    def tail:Tail

    def ::[A <: Base](a:A):HList[Base]
    def map[Out,F <: Func[Base,Out]](f:F):Map[Out,F]
  }

  case class HNil[Base]() extends HList[Base] {
    type Head = Nothing
    type Tail = Nothing
    type Map[Out,F <: Func[Base,Out]] = HNil[Out]

    def head = error("Head of an empty HList")
    def tail = error("Head of an empty HList")

    def ::[A <: Base](a:A) = HCons(a,this)
    def map[Out,F <: Func[Base,Out]](f:F) = new HNil[Out]
  }

  case class HCons[Base,A <: Base,B <: HList[Base]](head: A, tail: B) extends HList[Base] {
    type Head = A
    type Tail = B    
    type Map[Out,F <: Func[Base,Out]] = HCons[Out,F#Apply[Head],Tail#Map[Out,F]]

    def ::[C <: Base](c:C) = HCons(c,this)
    def map[Out,F <: Func[Base,Out]](f:F) =
      HCons(f(head),tail.map(f))
  }

  val :: = HCons 
}

object Test extends Application {
  import HList._

  val HNil = new HNil[Request[_]]

  val list = new Request[Int](1) :: new Request[String]("1") :: HNil

  val (a :: b :: HNil) = list
  val y:Request[String] = b

  val results = list.map[Any,Unwrap.type](Unwrap)

  val i:Int = results.head
}

import HList._
object Unwrap extends Func[Request[Any],Any] {
  type Apply[I <: Request[Any]] = I#O
  def apply[N <: Request[Any]](e:N) = null.asInstanceOf[Apply[N]]
}

另一个尝试是基于 Apocalisp 版本，它使用 fold 创建一个新的 HList，并再次产生了 Any 类型的 HList.任何提示将不胜感激.

The other attempt was based on the Apocalisp version which uses fold to create a new HList and again it resulted in a HList of Any types. Any tips would be appreciated.

推荐答案

HList 实现>shapeless 足够丰富以包含 HList 和 KList 功能.它提供了一个 map 操作，它应用一个更高等级的函数，可能具有特定类型的情况，跨它的元素产生一个适当类型的 HList 结果，

The HList implementation in shapeless is rich enough to subsume both HList and KList functionality. It provides a map operation which applies a higher-ranked function, possibly with type-specific cases, across it's elements yielding an appropriately typed HList result,

import shapeless.Poly._
import shapeless.HList._

// Define a higher-ranked function from Sets to Options
object choose extends (Set ~> Option) {
  def default[T](s : Set[T]) = s.headOption 
}

// An HList of Sets
val sets = Set(1) :: Set("foo") :: HNil

// Map our choose function across it ...
val opts = sets map choose

// The resulting value
opts == Option(1) :: Option("foo") :: HNil 

请注意，虽然在上面的例子中是这种情况，但并不要求 HList 元素共享一个公共的外部类型构造函数，它只是必须是这样的情况，映射到的高等级函数具有所有涉及的类型的情况,

Note that although it's the case in the above example there's no requirement that the HList elements share a common outer type constructor, it just has to be the case that the higher-ranked function mapped with has cases for all of the types involved,

// size is a higher-ranked function from values of arbitrary type to a 'size'
// which is defined as 1 by default but which has type specific cases for
// Strings and tuples
object size extends (Id ~> Const[Int]#λ) {
  def default[T](t : T) = 1
}
implicit def sizeString = size.λ[String](s => s.length)
implicit def sizeTuple[T, U](implicit st : size.λ[T], su : size.λ[U]) =
  size.λ[(T, U)](t => 1+size(t._1)+size(t._2))

size(23) == 1          // Default
size("foo") == 3       // Type specific case for Strings
size((23, "foo")) == 5 // Type specific case for tuples

现在让我们将其映射到 HList,

Now let's map this across an HList,

val l = 23 :: true :: "foo" :: ("bar", "wibble") :: HNil
val ls = l map size

ls == 1 :: 1 :: 3 :: 10 :: HNil

在这种情况下，被映射函数的结果类型是常量:无论参数类型是什么，它都是一个 Int.因此，生成的 HList 具有所有相同类型的元素，这意味着它可以有效地转换为普通列表，

In this case the result type of the function being mapped is constant: it's an Int no matter what the argument type is. Consequently the resulting HList has elements all of the same type, which means that it can usefully be converted to a vanilla list,

ls.toList == List(1, 1, 3, 10)

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