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
}
