映射無形HList的類型

Question

我一直在嘗試從scala的shapeless包中映射HList的類型，而無法訪問它們的值。

以下成功映射了HList的值

import shapeless._
import shapeless.Poly._
import ops.hlist.Mapper
import ops.hlist.Mapper._

trait Person {
  type Value
  val v : Value
}

case class StringPerson extends Person {
  type Value = String
  val v = "I like strings"
}

case class IntPerson extends Person {
  type Value = Int 
  val v = 42
}

object what_is_going_on {

  object test_value_op {
    val stringPerson = StringPerson()
    val intPerson = IntPerson()

    trait lpvfun extends Poly1 {
      implicit def default[A <: Person] = at[A](_.v)
    } 

    object vfun extends lpvfun {}

    // Use these to generate compiler errors if the mapped type is not what we'd expect:

    type TestListType = StringPerson :: IntPerson :: HNil
    type TestListExpectedMappedType = String :: Int :: HNil

    // Input:
    val testList : TestListType = stringPerson :: intPerson :: HNil

    // Output:
    val mappedList : TestListExpectedMappedType = testList map vfun

    // Get the actual mapped type 
    type TestListActualMappedType = mappedList.type

    // This compiles......
    val mappedList1 : TestListActualMappedType = mappedList

    // .... but weirdly this line doesn't. That isn't the point of this question, but I'd be very grateful for an answer.
    //implicitly[TestListActualMappedType =:= TestListExpectedMappedType]
  }

}

涼！ 除了由於某種原因不能implicitly[A =:= B]使用implicitly[A =:= B] ，還已經映射了HList的值，因此也映射了它們的類型。

現在，假設我們沒有HList值，但是我們知道它的類型。 我們如何映射其類型？

我根據此處 map的定義嘗試了以下方法：

object test_type_op { 
  type TestListType = StringPerson :: IntPerson :: HNil
  type TestListExpectedMappedType = String :: Int :: HNil

  // Attempt 1 does not work, compiler cannot prove =:=
  type MappedType = Mapper[vfun.type, TestListType]#Out
  implicitly[MappedType =:= TestListExpectedMappedType]

  // Attempt 2 does not work, compiler cannot prove =:=
  class GetMapper {
    implicit val mapper : Mapper[vfun.type, TestListType]
    implicitly[mapper.Out =:= TestListExpectedMappedType]
  }

}

如何在不訪問其值的情況下獲取映射的HList的類型？ 有沒有一種調試方法，為什么編譯器無法證明某些內容？ 謝謝您的閱讀。

Answer 1

在TestListActualMappedType的情況下，您獲得了mappedList的單例類型，這與推斷的mappedList類型mappedList 。 您可以看到完全相同的問題，而無需涉及Shapeless：

scala> val x = "foo"
x: String = foo

scala> implicitly[x.type =:= String]
<console>:13: error: Cannot prove that x.type =:= String.
       implicitly[x.type =:= String]
                 ^

您可以要求提供證據，證明x.type是String的子類型，或者可以使用shapeless.test.typed ，在您的情況下看起來像這樣：

import shapeless._, ops.hlist.Mapper

trait Person {
  type Value
  val v : Value
}

case class StringPerson() extends Person {
  type Value = String
  val v = "I like strings"
}

case class IntPerson() extends Person {
  type Value = Int 
  val v = 42
}

trait lpvfun extends Poly1 {
  implicit def default[A <: Person] = at[A](_.v)
} 

object vfun extends lpvfun {}

val stringPerson = StringPerson()
val intPerson = IntPerson()

val testList = stringPerson :: intPerson :: HNil
val mappedList = testList map vfun

shapeless.test.typed[String :: Int :: HNil](mappedList)

不過，這與真正地指定類型相比並沒有多大好處。

您可以要求提供證據，證明諸如Mapper類的類型類的輸出類型是您期望的特定輸入類型的類型：

scala> val m = Mapper[vfun.type, StringPerson :: IntPerson :: HNil]
m: shapeless.ops.hlist.Mapper[vfun.type,shapeless.::[StringPerson,shapeless.::[IntPerson,shapeless.HNil]]]{type Out = shapeless.::[String,shapeless.::[Int,shapeless.HNil]]} = shapeless.ops.hlist$Mapper$$anon$5@6f3598cd

scala> implicitly[m.Out =:= (String :: Int :: HNil)]
res1: =:=[m.Out,shapeless.::[String,shapeless.::[Int,shapeless.HNil]]] = <function1>

這更可能有用，但同樣取決於您要說服自己的確切方式。