Translate/encode Haskell's `data Obj = forall a. (Show a) => Obj a` 在 Scala 中

Question

我一直想不出如何在 Scala 中编码 Obj：

{-# LANGUAGE ExistentialQuantification #-}

data Obj = forall a. (Show a) => Obj a

instance Show Obj where show (Obj a) = "Obj " ++ show a

main = print $ show [Obj "hello", Obj 3, Obj True]

当运行时，以上生成以下输出：

[Obj "hello",Obj 3,Obj True]

然而，在 Scala 中，这似乎无法编译：

forSome { type T; implicit val ev: Show[T] }

这也不行：

forSome { type T : Show[T] }

这在类型系统级别是否可能，或者我是否需要 "capture" 类型 class 实例使用类似这样的东西：

class Obj[T](val x: T)(implicit val: Show[T])  // ...or similar

如有任何见解，我们将不胜感激！

Answer 1

你几乎答对了：

import scalaz._
import scalaz.Scalaz._

trait Obj {
  type T // existential type
  val x: T
  implicit val show: Show[T]
}

implicit val objSow: Show[Obj] = Show.shows[Obj] { (x: Obj) =>
  x.show.shows(x.x)
}

object Obj {
  /* "constructor" */
  def apply[U](_x: U)(implicit _show: Show[U]): Obj = new Obj {
    type T = U
    val x = _x
    val show = _show
  }
}

val test: List[Obj] = List(Obj(1), Obj(true), Obj("foo"))

/*
scala> test.shows
res0: String = [1,true,"foo"]
*/

P.S 我想在 apply 中使用 T 和 show；不是 U 和 _show。如果有人知道如何避免阴影，我将不胜感激！

或者您可以使用 forSome:

import scala.language.existentials

trait ObjE {
  val pair: Tuple2[T, Show[T]] forSome { type T }
}

/* And to define Show instance we have to help compiler unify `T` in pair components. */
def showDepPair[T] = Show.shows[Tuple2[T, Show[T]]] { x => x._2.shows(x._1) }
implicit val showObjE = Show.shows[ObjE] { x => showDepPair.shows(x.pair) }

这里要用Tuple2（或其他辅助类型）来捕获Show。我更喜欢以前的版本。对我来说，围绕类型成员更容易思考。

也在 Scala "Don Giovanni" forSome syntax will be eliminated in favour of val pair: ({ type λ[T] = Tuple2[T, Show[T]] })#λ[_] }, which works already too. I hope there will be some syntax support for type lambdas as well. kind-projector doesn't help in this situation (repeated use of T). Maybe something like in Typelevel scalac 中：val pair: ([T] => Tuple2[T, Show[T])[_])。

另一个根本性的变化是：

A single fundamental concept – type members – can give a precise meaning to generics, existential types, wildcards, and higher-kinded types.

所以从编译器的角度来看，这两种形式是等价的（以前我们解包元组）。 我不是 100% 确定目前有什么区别，如果有的话。

P.S。 The Troubles with Types 帮助我理解了 scala 当前类型系统的怪癖。

Answer 2

我已经 "packaged" Oleg 对这个通用且（看似）可重用结构的回答：

import scala.language.{ higherKinds, implicitConversions }

trait AnyWithTC[TC[_]] { type T; val x: T; implicit val ev: TC[T] }

// don't like the 'implicit' here; suggestions welcome
implicit def AnyWithTC[T, TC[_]](x: T)(implicit ev: TC[T]) = {
  type T0 = T; val x0 = x; val ev0 = ev
  new AnyWithTC[TC] { type T = T0; val x = x0; val ev = ev0 }
}

那么，data Obj = forall a. (Show a) => Obj a可以这样实现：

type Obj = AnyWithTC[Show]
implicit val objShow = Show.shows[Obj] { x => "Obj " + x.show.shows(x.x)   }
val xs: List[Obj] = List(1, true, "hello")
println(xs.shows) // prints [Obj 1,Obj true, Obj hello]

Translate/encode Haskell's `data Obj = forall a. (Show a) => Obj a` 在 Scala 中

Translate/encode Haskell's `data Obj = forall a. (Show a) => Obj a` in Scala

scala

haskell

typeclass

implicit

existential-type