奇怪的Scala错误

发布时间：2020-12-16 18:07:16 所属栏目：安全来源：网络整理

导读：我试图创建基于抽象回合的游戏和抽象AI： abstract class AGame { type Player type Move // Player inside def actPlayer : Player def moves (player : Player) : Iterator[Move] def play (move : Move) def undo () def isFinished : Boolean def result

我试图创建基于抽象回合的游戏和抽象AI：

abstract class AGame {
  type Player
  type Move     // Player inside

  def actPlayer : Player
  def moves (player : Player) : Iterator[Move]
  def play (move : Move)
  def undo ()
  def isFinished : Boolean
  def result (player : Player) : Double
}

abstract class Ai[Game <: AGame] {
  def genMove (player : Game#Player) : Game#Move
}

class DummyGame extends AGame {
  type Player = Unit
  type Move = Unit

  def moves (player : Player) = new Iterator[Move] {
    def hasNext = false
    def next = throw new Exception ("asd")
  }

  def actPlayer = ()

  def play (move : Move) {
  }

  def undo () {
  }

  def isFinished = true

  def result (player : Player) = 0
}

class DummyAi[Game <: AGame] (game : Game) extends Ai[Game] {
  override def genMove (player : Game#Player) : Game#Move = {
    game.moves (player).next
  }
}

我认为我必须使用像Game#Player这样奇怪的类型访问器.
我得到了非常令人费解的错误.我想了解一下：

[error] /home/lew/Devel/CGSearch/src/main/scala/Main.scala:41: type mismatch;
[error]  found   : Game#Player
[error]  required: DummyAi.this.game.Player
[error]     game.moves (player).next
[error]                 ^

解决方法

def移动(玩家：玩家)意味着移动接受该游戏的玩家.

Game#Player是任何游戏玩家的类型.所以移动(玩家)是一种类型不匹配.

这是一个简单的例子,说明为什么它必须是不匹配的.假设它不是,看看接下来会发生什么：

class Game2 extends DummyGame {
  override type Player = Boolean
  override type Move = Boolean

  override def moves(player : Boolean) = new Iterator[Boolean] {...}
}

val game2: DummyGame = new Game2
// game2.Player is Boolean

val dummyGameAi = new DummyAi[DummyGame](game2)
// DummyGame#Player == Unit,so the type of genMove for Ai[DummyGame] is
// def genMove (player : Unit) : Unit

dummyGameAi.genMove(())
// this calls game2.moves(()),which doesn't typecheck

为了实现这一点,我们可以更改genMove的类型.如果我们将游戏作为参数传递(并且它仍然有意义),我们可以使用路径依赖类型：

abstract class Ai[Game <: AGame] {
  def genMove (game : Game)(player : game.Player) : game.Move
  // now game.moves (player) typechecks
}

（编辑：李大同）

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