斯卡拉OO-可选扩展特性

Question

不知道这样的事情是否甚至可能，但我期望达到的效果如下： 我有一个被称为物理位置的特性，它需要一个int（对应一个人的城市）作为子类的变量存在。

class Building extends PhysicallySituated //A building has an address 
class OilTanker extends Ship //An oiltanker is always moving around 
class Yacht extends Ship with Option[ PhysicallySituated ] //a Yacht could have an address 

我该如何模拟这种情况，因为PhysicallySituated具有在位置上操作的方法？我在寻找的是调用船上的方法返回一个可选的值，它是PhysicallySituated中的原始方法返回的类型。我知道我可以使用PhysicallySituated的一个实例并调用这个实例的方法，但是这违背了封装的最佳实践。

我要指出，我的斯卡拉的理解是中等水平，但我愿意，如果他们指出

Answer 1

会为你的工作读了任何先进的理念？ （RAN使用Scala REPL）：

trait Address 

case object EmptyAddress extends Address 

case class NonEmptyAddress(addrStr: String) extends Address 

trait Situated { 
    val address: Address 
} 

trait OptionallySituated extends Situated { 
    val address = EmptyAddress 
} 

trait PhysicallySituated extends Situated { 
    val address: NonEmptyAddress 
} 

class Building extends PhysicallySituated { 
    val address = NonEmptyAddress("this is my place") 
} 

class Ship 

class OilTanker extends Ship 

class Yacht extends Ship with OptionallySituated 

scala> val stuff = List(new Building, new Yacht) 
stuff: List[Situated{val address: Product with Serializable with Address}] = List(Building@7e1f584d, Yacht@7dff6d05) 

scala> stuff.map(_.address) 
res0: List[Product with Serializable with Address] = List(NonEmptyAddress(this is my place), EmptyAddress)       

您可以使用Option[Address]与Some和None而不是定义上面以及地址类。