scala - 为什么我不能索引到通过映射另一个 HList 获得的 HList？

Question

甚至.head不起作用。

我需要做哪些改变才能完成这项工作？

import shapeless._
import HList._
import Nat._

scala> case class Foo[A](a: A)
defined class Foo

scala> case class Bar[A](f: Foo[A])
defined class Bar

scala> val xs = Foo(23) :: Foo("blah") :: HNil
xs: shapeless.::[Foo[Int],shapeless.::[Foo[java.lang.String],shapeless.HNil]] = Foo(23) :: Foo(blah) :: HNil

scala> object mapper extends (Foo ~> Bar) {
     |   def apply[A](f: Foo[A]) = Bar(f)
     | }
defined module mapper

scala> xs map mapper
res13: mapper.Out = Bar(Foo(23)) :: Bar(Foo(blah)) :: HNil

scala> res13.apply[_1]
<console>:38: error: could not find implicit value for parameter at: shapeless.At[mapper.Out,shapeless.Nat._1]
              res13.apply[_1]
                         ^

scala> res13.head
<console>:38: error: could not find implicit value for parameter c: shapeless.IsHCons[mapper.Out]
              res13.head
                    ^

Answer 1

至少在最新的 2.10.0-SNAPSHOT 中完全符合我的要求，

import shapeless._
import HList._
import Nat._

scala> case class Foo[A](a: A)
defined class Foo

scala> case class Bar[A](f: Foo[A])
defined class Bar

scala> val xs = Foo(23) :: Foo("blah") :: HNil
xs: shapeless.::[Foo[Int],shapeless.::[Foo[String],shapeless.HNil]] = Foo(23) :: Foo(blah) :: HNil                                                                                                     

scala> object mapper extends (Foo ~> Bar) {
     |   def apply[A](f: Foo[A]) = Bar(f)                                                                                                                                                              
     | }                                                                                                                                                                                               
defined module mapper                                                                                                                                                                                  

scala> xs map mapper                                                                                                                                                                                   
res0: shapeless.::[Bar[Int],shapeless.::[Bar[String],shapeless.HNil]] = Bar(Foo(23)) :: Bar(Foo(blah)) :: HNil                                                                                         

scala> res0[_1]
res1: Bar[String] = Bar(Foo(blah))                                                                                                                                                                     

scala> res0.head                                                                                                                                                                                       
res2: Bar[Int] = Bar(Foo(23))

res0请注意，上面推断的类型Bar[Int] :: Bar[String] :: HNil不是mapper.Out......我怀疑这是 2.9.x 和 2.10.0-SNAPSHOT 之间的行为差​​异。

如果您坚持使用 2.9.x，那么我认为您应该能够通过明确地归因Bar[Int] :: Bar[String] :: HNil于您来解决该问题res13......显然这更冗长，但 c'est la vie。