18

Given a Shapeless HList where every list element shares the same type constructor, how can the HList be sequenced?

For example:

def some[A](a: A): Option[A] = Some(a)
def none[A]: Option[A] = None

val x = some(1) :: some("test") :: some(true) :: HNil
val y = sequence(x) // should be some(1 :: "test" :: true :: HNil)

def sequence[L <: HList : *->*[Option]#λ, M <: HList](l: L): Option[M] =
  ???

I tried to implement sequence like this:

object optionFolder extends Poly2 {
  implicit def caseOptionValueHList[A, B <: HList] = at[Option[A], Option[B]] { (a, b) =>
    for { aa <- a; bb <- b } yield aa :: bb
  }
}

def sequence[L <: HList : *->*[Option]#λ, M <: HList](l: L): Option[M] = {
  l.foldRight(some(HNil))(optionFolder)
}

But that does not compile:

could not find implicit value for parameter folder: shapeless.RightFolder[L,Option[shapeless.HNil.type],SequencingHList.optionFolder.type]

Any tips on implementing this for either a specific example like Option or for an arbitrary Applicative?

4

2 回答 2

18

你非常接近 - 你只需要确保你有它要求的额外证据：

def sequence[L <: HList : *->*[Option]#λ, M <: HList](l: L)(implicit
  folder: RightFolder[L, Option[HNil], optionFolder.type]
) = l.foldRight(some(HNil: HNil))(optionFolder)

或者，如果你想要更通用的东西，并且有一个这样的应用实现：

trait Applicative[F[_]] {
  def ap[A, B](fa: => F[A])(f: => F[A => B]): F[B]
  def point[A](a: => A): F[A]
  def map[A, B](fa: F[A])(f: A => B): F[B] = ap(fa)(point(f))
}

implicit object optionApplicative extends Applicative[Option] {
  def ap[A, B](fa: => Option[A])(f: => Option[A => B]) = f.flatMap(fa.map)
  def point[A](a: => A) = Option(a)
}

你可以写：

object applicativeFolder extends Poly2 {
  implicit def caseApplicative[A, B <: HList, F[_]](implicit
    app: Applicative[F]
  ) = at[F[A], F[B]] {
    (a, b) => app.ap(a)(app.map(b)(bb => (_: A) :: bb))
  }
}

def sequence[F[_]: Applicative, L <: HList: *->*[F]#λ, M <: HList](l: L)(implicit
  folder: RightFolder[L, F[HNil], applicativeFolder.type]
) = l.foldRight(implicitly[Applicative[F]].point(HNil: HNil))(applicativeFolder)

现在您也可以序列列表等（假设您有适当的实例）。

更新：请注意，我sequence在这两种情况下都省略了返回类型注释。如果我们把它放回去，编译器就会窒息：

<console>:18: error: type mismatch;
 found   : folder.Out
 required: F[M]

这是因为RightFolder实例携带其返回类型作为抽象类型成员。我们知道F[M]在这种情况下，但编译器并不关心我们知道什么。

如果我们希望能够明确返回类型，我们可以使用RightFolderAux实例来代替：

def sequence[F[_]: Applicative, L <: HList: *->*[F]#λ, M <: HList](l: L)(implicit
  folder: RightFolderAux[L, F[HNil], applicativeFolder.type, F[M]]
): F[M] =
  l.foldRight(implicitly[Applicative[F]].point(HNil: HNil))(applicativeFolder)

请注意，它RightFolderAux有一个额外的类型参数，它指示返回类型。

于 2013-04-21T04:45:18.410 回答
1

Now you can use kittens cats.sequence

import cats.implicits._
import cats.sequence._
import shapeless._

val f1 = (_: String).length
val f2 = (_: String).reverse
val f3 = (_: String).toDouble

val f = (f1 :: f2 :: f3 :: HNil).sequence
assert( f("42.0") == 4 :: "0.24" :: 42.0 :: HNil)

The example sequenced against Function but you can use anything that has an cats Applicative instance.

于 2016-01-14T20:40:03.620 回答