如何使类型类与Scala中的异构列表一起使用 [英] How to make a typeclass works with an heterogenous List in scala

问题描述

给出以下类型类和一些常见类型的实例

Given the following typeclass and some instances for common types

trait Encoder[A] {
  def encode(a: A): String
}

object Encoder {
  implicit val stringEncoder = new Encoder[String] {
    override def encode(a: String): String = a
  }
  implicit val intEncoder = new Encoder[Int] {
    override def encode(a: Int): String = String.valueOf(a)
  }
  implicit def listEncoder[A: Encoder] =
    new Encoder[List[A]] {
      override def encode(a: List[A]): String = {
        val encoder = implicitly[Encoder[A]]
        a.map(encoder.encode).mkString(",")
      }
    }
}

有没有办法使它起作用

Encoder.listEncoder.encode(List("a", 1))

推荐答案

如果为 Any

object Encoder {
  implicit val anyEncoder = new Encoder[Any] {
    override def encode(a: Any): String = a.toString
  }

  //instances for String, Int, List[A]
}

然后

Encoder.listEncoder[Any].encode(List("a", 1))

将起作用.

您必须在此处明确指定类型参数( listEncoder [Any] )，因为这就是在scala中类型推断的工作方式.确实，在推断出 encode 的参数类型

You have to specify type parameter here explicitly (listEncoder[Any]) because that's how type inference work in scala. Indeed, after the type of argument of encode is inferred

Encoder.listEncoder[???].encode(List("a", 1))
//                              ^^^^^^^^^^^^
//                              List[Any]

现在回来推断 listEncoder 的类型参数为时已晚.

it's too late to come back to infer the type parameter of listEncoder.

如果定义了委托人功能

def encode[A](a: A)(implicit encoder: Encoder[A]): String = encoder.encode(a)

或语法

implicit class EncoderOps[A](a: A)(implicit encoder: Encoder[A]) {
  def encode: String = encoder.encode(a)
}

    // or
// implicit class EncoderOps[A](a: A) {
//   def encode(implicit encoder: Encoder[A]): String = encoder.encode(a)
// }

那么您不必显式指定类型参数

then you don't have to specify type parameter explicitly

encode("a")
encode(1)
encode(List("a", 1))

"a".encode
1.encode
List("a", 1).encode

顺便说一句， List("a"，1)不是异类列表，它是元素类型为 Any 的普通同质列表.异构列表将是

By the way, List("a", 1) is not a heterogenous list, it's an ordinary homogenous list with element type Any. A heterogenous list would be

val l: String :: Int :: HNil = "a" :: 1 :: HNil

---

您可以尝试磁铁，而不是或 HList 而不是 List [A]

sealed trait HList 
case class ::[+H, +T <: HList](head: H, tail: T) extends HList 
case object HNil extends HList 
type HNil = HNil.type

implicit class HListOps[L <: HList](l: L) {
  def ::[A](a: A): A :: L = new ::(a, l)
}

trait Encoder[A] {
  def encode(a: A): String
}

object Encoder {
  implicit val stringEncoder: Encoder[String] = new Encoder[String] {
    override def encode(a: String): String = a
  }
  implicit val intEncoder: Encoder[Int] = new Encoder[Int] {
    override def encode(a: Int): String = String.valueOf(a)
  }
  implicit val hnilEncoder: Encoder[HNil] = new Encoder[HNil] {
    override def encode(a: HNil): String = ""
  }
  implicit def hconsEncoder[H, T <: HList](implicit
                                           hEncoder: Encoder[H],
                                           tEncoder: Encoder[T]
                                          ): Encoder[H :: T] = new Encoder[H :: T] {
    override def encode(a: H :: T): String = 
      s"${hEncoder.encode(a.head)},${tEncoder.encode(a.tail)}"
  }
}

def encode[A](a: A)(implicit encoder: Encoder[A]): String = encoder.encode(a)

implicit class EncoderOps[A](a: A)(implicit encoder: Encoder[A]) {
  def encode: String = encoder.encode(a)
}

val l: String :: Int :: HNil = "a" :: 1 :: HNil
encode(l)
l.encode

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