我想采用这个模式:
def accept[T](a: RList[T]) = true
def accept[T, V](a: RList[T], b: RList[V])(implicit ev: a.S =:= b.S) = true
def accept[T, V, Q](a: RList[T], b: RList[V], c: RList[Q])(implicit ev: a.S =:= b.S, ev2: b.S =:= c.S) = true但是让它接受一个KList,而不是手动覆盖所有的奇偶校验。基本上,我想说,“使用任意数量的RList,它们具有相同的S成员类型”
RList是一个包含S类型的特征。(有关RList的更多背景信息以及我为什么要这样做,请参阅:Constrain function based on origin (Path Dependent type? Type Generation?))
发布于 2012-08-21 14:09:54
看起来您所做的只是试图让编译器检查类型是否一致,因为您的方法总是返回true。
你能不能接受一个“RLists列表”,保证S个都匹配,而不是让方法接受任意数量的匹配?
下面是这样一个列表的构建方式:
package rl {
// A simplified version of your RList:
trait RList[T] {
type S
def data: List[T]
}
// A list of RLists which have identical S
sealed trait RListList
// RListNil is an empty list
trait RListNil extends RListList {
def ::[H <: RList[_]](h: H) = rl.::[h.S,H,RListNil](h, this)
}
// there is exactly one RListNil
case object RListNil extends RListNil
// List can be a cons cell of lists sharing the same S
final case class ::[S, H <: RList[_], T <: RListList](head: H, tail: T) extends RListList {
// We only allow you to cons another to this if we can find evidence that the S matches
def ::[H2 <: RList[_]](h: H2)(implicit ev: =:=[h.S,S]) = rl.::[S,H2,::[S,H,T]](h, this)
}现在,如果我们试图构造一个不是所有S类型都一致的RListList,编译器将捕获我们:
object RListTest {
val list1 = new RList[Int] { type S = String; def data = List(1,2,3,4) }
val list2 = new RList[String] { type S = String; def data = List("1","2","3","4") }
val list3 = new RList[Double] { type S = Float; def data = List(1.1,2.2,3.3,4.4) }
val listOfLists1 = list1 :: RListNil // fine
val listOfLists2 = list2 :: listOfLists1 // still fine, since list1 and list2 have the same S
val listOfLists3 = list3 :: listOfLists2 // compiler error: Cannot prove that java.lang.String =:= Float
}这是使用依赖方法类型,这意味着您需要使用scala 2.10,或者需要在2.9.x中使用-Y依赖方法类型开关进行编译
https://stackoverflow.com/questions/12046974
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