如何使用无形状类型为具有依赖类型的类型创建实例

Question

我试图为具有依赖类型的类型类派生一个元组实例。我使用“无形”来为元组元素创建调用类型类。我很难匹配元组实例类型：

import shapeless.the
import simulacrum.typeclass

@typeclass trait Identifiable[M] {
  type K
  def identify(id: M): K
}

object Identifiable{
  implicit def identifiableTuple[K1: Identifiable, K2: Identifiable]: Identifiable[(K1,K2)] = new Identifiable[(K1,K2)]{
     val b = the[Identifiable[K2]]
    val a = the[Identifiable[K1]]
    type K = (a.K, b.K)   
    override def identify(id: (K1, K2)): K = {
          val k1 = the[Identifiable[K1]].identify(id._1)
          val k2 = the[Identifiable[K2]].identify(id._2)
          (k1,k2)
        }
  }

我知道这个错误：

type mismatch;
 found   : k1.type (with underlying type ai.fugo.cms.service.common.domain.Identifiable[K2]#K)
 required: this.a.K

type mismatch;
 found   : k2.type (with underlying type ai.fugo.cms.service.common.domain.Identifiable[K1]#K)
 required: this.b.K

Answer 1

尝试Aux模式

trait Identifiable[M] {
  type K
  def identify(id: M): K
}

object Identifiable {
  type Aux[M, K0] = Identifiable[M] { type K = K0 }

  implicit def identifiableTuple[M1, K1, M2, K2](
    implicit
    identifiable1: Identifiable.Aux[M1, K1],
    identifiable2: Identifiable.Aux[M2, K2]
  ): Identifiable.Aux[(M1, M2), (K1, K1)] = new Identifiable[(M1, M2)] {
    type K = (K1, K2)
    def identify(id: (M1, M2)): (K1, K2) =
      identifiable1.identify(id._1) -> identifiable2.identify(id._2)
  }
}

Aux模式的发明是因为

，

• ，人类更容易阅读
• ，我认为(？)编译器过去在派生路径相关类型的类型类时遇到了问题.但不是因为他们的化名

所以，只需使用Aux来派生事物。