1

假设我有这样的功能：

val fooXAB: X => A => Try[B] = ...
val fooXBC: X => B => Try[C] = ...
val fooXCD: X => C => Try[D] = ...

我想将它们组合成一个新函数，它依次fooXAD: X => A => Try[D]调用fooXAB、fooXBC和，并将参数传递给它们。fooXCDX

假设我使用scalaz并有一个 monad 实例scala.util.Try。现在我可以这样做：

type AB = Kleisli[Try, A, B]
type BC = Kleilsi[Try, B, C]
type CD = Kleisli[Try, C, D]

type XReader[T] = Reader[X, T]

val fooXAB: XReader[AB] = ...
val fooXBC: XReader[BC] = ...
val fooXCD: XReader[CD] = ...

val fooXAC: XReader[AC] =
  for {
    ab <- fooXAB
    bc <- fooXBC
    cd <- fooXCD
  } yield (ab andThen bc andThen cd)

是否有意义？是否可以简化它？

4

1 回答 1

1

所以我认为外部函数上的 Reader Monad 在这里没有帮助。一旦你将和 X 应用到你的三个XReader，你就可以对结果使用Kleisli组合（假设你有一个用于 Try 的 Monad 实例）。这是您以这种方式重新编写的示例，它为我编译：

import scala.util.{Try,Success}

import scalaz._
import Scalaz._

object A
object B
object C
object D
trait X

object Main {
  implicit val pretendTryIsAMonad: Monad[Try] = new Monad[Try] {
    def point[A](a: => A): Try[A] = Success(a)
    def bind[A,B](fa: Try[A])(f: A => Try[B]): Try[B] = fa flatMap f
  }

  type AB = Kleisli[Try, A.type, B.type]
  type BC = Kleisli[Try, B.type, C.type]
  type CD = Kleisli[Try, C.type, D.type]
  type AD = Kleisli[Try, A.type, D.type]

  type XReader[T] = X => T

  val fooXAB: XReader[AB] = (x: X) => Kleisli((a: A.type) => Success(B))
  val fooXBC: XReader[BC] = (x: X) => Kleisli((b: B.type) => Success(C))
  val fooXCD: XReader[CD] = (x: X) => Kleisli((c: C.type) => Success(D))

  val fooXAD: XReader[AD] = (x: X) =>
    fooXAB(x) >=> fooXBC(x) >=> fooXCD(x)
}
于 2016-01-12T17:03:55.747 回答