scala - 简单的 Shapeless HLIST 上的模式匹配

Question

我使用 shapeless 库编写了这个简单的代码

import shapeless.LabelledGeneric
case class Icecream(name: String, numberOfCherries: Int, inCone: Boolean)

object ShapelessRecordEx2 extends App {
   val gen = LabelledGeneric[Icecream]
   val hlist = gen.to(Icecream("vanilla", 2, false))
   hlist match {
      case h :: _ => println(h)
   }
}

但甚至不编译

Error:(12, 14) constructor cannot be instantiated to expected type;
 found   : scala.collection.immutable.::[B]
 required: shapeless.::[String with shapeless.labelled.KeyTag[Symbol with shapeless.tag.Tagged[String("name")],String],shapeless.::[Int with shapeless.labelled.KeyTag[Symbol with shapeless.tag.Tagged[String("numberOfCherries")],Int],shapeless.::[Boolean with shapeless.labelled.KeyTag[Symbol with shapeless.tag.Tagged[String("inCone")],Boolean],shapeless.HNil]]]
      case h :: _ => println(h)

如果我使用的是普通列表，这段代码就可以了。

Answer 1

您只需要导入，默认情况下 scala.Predef进口 ::运算符(operator)来自 scala.collection.immutable.List .

import shapeless.LabelledGeneric
import shapeless.::
case class Icecream(name: String, numberOfCherries: Int, inCone: Boolean)

object ShapelessRecordEx2 extends App {
   val gen = LabelledGeneric[Icecream]
   val hlist = gen.to(Icecream("vanilla", 2, false))
   hlist match {
      case h :: _ => println(h)
   }
}

还有另一种选择，导入 ListCompat._ .

import shapeless.HList.ListCompat._

object ShapelessRecordEx2 extends App {
  val gen = LabelledGeneric[Icecream]
  val hlist = gen.to(Icecream("vanilla", 2, false))
  hlist match {
    case h #: _ => println(h)
  }
}