Scala ADT类型不匹配

Question

我对定义endState的行的编译错误感到困惑。这是代码：

import java.util.UUID
object Evaluation {
  def default: Evaluation[Unit, Unit] = new Evaluation[Unit, Unit](identity)
}
case class Evaluation[From, +To](evaluate: (From) => To)
object FSMLike {
  val uuidEmpty: UUID = new UUID(0L, 0L)
  val endState: Evaluation[Unit, FSMLike[Nothing]] = new Evaluation(() => FSMEmpty)
  lazy val stop: FSMEntry[Unit, Unit, Nothing] = FSMEntry(uuidEmpty, Evaluation.default, endState)
  def apply[From1, From2, To](
    action: Evaluation[From1, Unit],
    nextState: Evaluation[From2, FSMLike[To]]
  ): (UUID, FSMLike[To]) = {
    val uuid = UUID.randomUUID
    uuid -> FSMEntry(uuid, action, nextState)
  }
}
sealed trait FSMLike[+A]
case object FSMEmpty extends FSMLike[Nothing]
case class FSMEntry[From1, From2, +To](
  id: UUID,
  action: Evaluation[From1, Unit],
  nextState: Evaluation[From2, FSMLike[To]]
) extends FSMLike[To] {
  def transition(arg1: From1, arg2: From2): FSMLike[To] = {
    action.evaluate(arg1)
    nextState.evaluate(arg2)
  }
  override def toString: String = s"^$id^"
}

这是错误：

Error:(14, 72) type mismatch;
 found   : () => FSMEmpty.type (with underlying type () => FSMEmpty.type)
 required: Unit => FSMLike[Nothing]
  val endState: Evaluation[Unit, FSMLike[Nothing]] = new Evaluation(() => FSMEmpty)

Answer 1

您正在尝试通过 () => FSMEmpty，即没有参数的函数，其中一个具有一个类型Unit的函数。当然，当您使用()作为表达式时，它是Unit类型的唯一值，但是=>的左侧不是表达式。

您应该写_ => FSMEmpty。我认为{ case () => FSMEmpty }也可以正常工作，但是使用它没有太多意义。