Why does mapping over an HList of Option[T] not work?

问题

This does not compile and I do not understand why:

import shapeless._
import poly._

object option extends (Option ~> List) {
    def apply[T](t: Option[T]) = t.toList
}

val simple = Some(1) :: Some("hello") :: Some(true) :: HNil
val opts: List[Int] :: List[String] :: List[Boolean] :: HNil = simple map option

This is somewhat surprising, because this does compile:

import shapeless._
import poly._

object choose extends (Set ~> Option) {
    def apply[T](s: Set[T]) = s.headOption
}

val sets = Set(1) :: Set("foo") :: HNil
val opts: Option[Int] :: Option[String] :: HNil = sets map choose

回答1:

The problem is your usage of Some instead of Option. The compiler complains that it can't find anything to convert a list of Somes though your Option mapper is being used. To fix this you can:

Change your simple list construction to use Option instead of Some:

  object option extends (Option ~> List) {
    def apply[T](t: Option[T]) = t.toList
  }

  val simple = Option(1) :: Option("hello") :: Option(true) :: HNil
  val opts2: List[Int] :: List[String] :: List[Boolean] :: HNil = simple map option

Use Some as the type of your mapper:

  object option extends (Some ~> List) {
    def apply[T](t: Some[T]) = t.toList
  }

  val simple = Some(1) :: Some("hello") :: Some(true) :: HNil
  val opts2: List[Int] :: List[String] :: List[Boolean] :: HNil = simple map option

Or force the type of simple to be downcast to Option:

  object option extends (Option ~> List) {
    def apply[T](t: Option[T]) = t.toList
  }

  val simple: Option[Int] :: Option[String] :: Option[Boolean] :: HNil = Some(1) :: Some("hello") :: Some(true) :: HNil
  val opts2: List[Int] :: List[String] :: List[Boolean] :: HNil = simple map option

来源：https://stackoverflow.com/questions/26127542/why-does-mapping-over-an-hlist-of-optiont-not-work

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scala

mapping

shapeless

hlist