类类型作为在Scala中的地图中的键 [英] Class type as key in map in Scala

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本文介绍了类类型作为在Scala中的地图中的键的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我的游戏有

  class Enemy 

谁是AI /功能我可以改变

  trait移动
trait VerticalMover扩展移动
trait RandomMover扩展移动

等。现在我需要基于trait获取预加载的东西。我想做的是有一个地图,接受所有的traits,扩展移动作为键,然后一些EnemyContainer作为值将具有trait相关内容预加载。



但是如何我定义这样的地图,以及如何格式化我的.get()来获取容器的一些敌人的实例。像:

  val myEnemy = new Enemy with RandomMover 
val myDetails:EnemyContainer = enemyDetailsS​​tore.get(myEnemy.getClass )


解决方案

Map [Class [_< ;: Moving],EnemyDetails] 类型。我怀疑像:

  //给出一个Map [Class [_< ;: Moving],EnemyDetails]键
enemyDetailsS​​tore.filterKeys(_ isInstance myEnemy)

或:

  // Iterable [EnemyDetails] 
enemyDetailsS​​tore collect {case(c,d)if c isInstance myEnemy => d}

或者只是:

  //选项[EnemyDetails] 
enemyDetailsS​​tore collectFirst {case(c,d)if c isInstance myEnemy => d}

会为你做。这个代码的唯一问题是它是O(N),因为它需要一个遍历的地图,而不是简单的查找,这将是O(1)或O(log N)


My game has

class Enemy

who's AI/functionality I can change with

trait Moving
trait VerticalMover extends Moving
trait RandomMover extends Moving

and so on. Now I need to fetch preloaded stuff based on trait. What I would like to do is have a Map that accepts all traits that extend Moving as keys and then some EnemyContainer as value that would have trait related content preloaded.

But how do I define such a Map and how do format my .get() to get the container by an instance of some Enemy. Something like:

val myEnemy = new Enemy with RandomMover 
val myDetails:EnemyContainer = enemyDetailsStore.get(myEnemy.getClass)

解决方案

Well, I assume that your enemy details store is of type Map[Class[_ <: Moving], EnemyDetails]. I suspect that something like:

//gives a Map[Class[_ <: Moving], EnemyDetails] for all matching keys
enemyDetailsStore.filterKeys(_ isInstance myEnemy) 

Or:

//Iterable[EnemyDetails]
enemyDetailsStore collect { case (c, d) if c isInstance myEnemy => d }

Or even just:

//Option[EnemyDetails]
enemyDetailsStore collectFirst { case (c, d) if c isInstance myEnemy => d }

Will do for you. The only "issue" with this code is that it's O(N), in that it requires a traversal of the map, rather than a simple lookup, which would be O(1), or O(log N)

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