Scala 类型类和隐式谜题

Question

下面的最后一行语句产生错误： “类型不匹配；发现：TestExpandableWithLibrary.this.library.type（具有基础类型 org.typeclass.Library）需要：V”

这是我尝试进行隐式转换的地方。我简单地使用 typeclass 函数的前一行确实可以正常工作。

关于如何解决它的任何想法？

package org.typeclass

///////////////////////////////////////////////////////////////////////////////
// the domain objects

case class Book(bookName: String)

case class Library(libraryName: String, books: Set[Book])

object Library {
  def apply(libraryName: String, bookNames: String*): Library =
    Library(libraryName, bookNames.map(Book(_)).toSet)
}

case class TreeClass(nodeName: String, children: Seq[TreeClass])

///////////////////////////////////////////////////////////////////////////////
// the typeclass definition

trait Expandable[T, V, R] {
  def expandWith(template: T, values: V): R
}

object Expandable {

  def expandWithF[T, V, R](template: T, values: V)(implicit ev: Expandable[T, V, R]): R =
    ev.expandWith(template, values)

  implicit class ExpandableItem[T, V, R](val template: T) extends AnyVal {
    def expandWithM(values: V)(implicit ev: Expandable[T, V, R]): R =
      ev.expandWith(template, values)
  }
}

///////////////////////////////////////////////////////////////////////////////
// a typeclass implementation

object ExpandableImpls {

  implicit object ExpandableTreeClass extends Expandable[TreeClass, Library, TreeClass] {
    def expandWith(template: TreeClass, library: Library): TreeClass = {
      val parentName = s"${template.nodeName}.${library.libraryName}"
      val children = library.books.map(book => TreeClass(s"${parentName}.${book.bookName}", Seq.empty)).toSeq
      TreeClass(parentName, children)
    }
  }

}

//@RunWith(classOf[JUnitRunner])
class TestExpandableWithLibrary /*extends FlatSpec*/ {

  import Expandable._
  import ExpandableImpls._

  val library = Library("test", "black", "white")
  val root = TreeClass("parent", Seq.empty)

  val useF = expandWithF(root, library) // this works

  val useM = root.expandWithM(library) // this doesn't work!

}

Answer 1

问题在于您需要将后两个类型参数放在扩展方法上的隐式类上——因为它们将被推断为Nothing，因为它们在构造函数中没有被引用论据。

implicit class ExpandableItem[T](val template: T) extends AnyVal {
  def expandWithM[V, R](values: V)(implicit ev: Expandable[T, V, R]): R =
    ev.expandWith(template, values)
}

这应该可以正常工作。