[英]Common supertype of methods
请考虑以下定义:
trait Event
case class Event1[A] extends Event
case class Event2[A, B] extends Event
/* ... */
trait Filter { val cond: Event => Boolean }
case class Filter1[A](cond: Event1[A] => Boolean) extends Filter
case class Filter2[A, B](cond: Event2[A, B] => Boolean) extends Filter
/* ... */
我想我在这里要完成的工作很清楚:我想确保每次遇到Filter
,都保证有一个cond
函数,该函数接受Event
的相应子类型并给我一个布尔值。 显然,上面的代码无法编译,例如, Event1[A] => Boolean
实际上不是Event => Boolean
的子类型。 一个人将如何解决这一问题?
如下所示呢?
sealed trait Event
case class Event1[A]() extends Event
case class Event2[A, B]() extends Event
/* ... */
sealed trait Filter[T <: Event] { val cond: T => Boolean }
case class Filter1[A](cond: Event1[A] => Boolean) extends Filter[Event1[A]]
case class Filter2[A, B](cond: Event2[A, B] => Boolean) extends Filter[Event2[A, B]]
或者,您可以覆盖抽象类型,而不是使用参数化类型:
sealed trait Filter {
type Filterable
val cond: Filterable => Boolean
}
case class Filter1[A](cond : Event1[A] => Boolean) extends Filter{
override type Filterable = Event1[A]
}
case class Filter2[A, B](cond: Event2[A, B] => Boolean) extends Filter{
override type Filterable = Event2[A, B]
}
尝试这个:
trait Event
case class Event1[A](a: A) extends Event
case class Event2[A, B](a: A, b: B) extends Event
trait Filter[T <: Event] { val cond: T => Boolean }
case class Filter1[A](cond: Event1[A] => Boolean) extends Filter[Event1[A]]
case class Filter2[A, B](cond: Event2[A, B] => Boolean) extends Filter[Event2[A, B]]
它为我编译
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