假设我们有一个T[A,B]
类型,我们可以表达以下类型约束,让我们称之为HT
:
满足HT
的每种类型都必须是T[P1, P2] :: T[P2, P3] :: T[P3, P4] :: ... :: T[PN-1, PN] :: HNil
type x = P1 :: P2 :: ... :: PN :: HNil
。
我试图在类型化的顺序处理管道上找到一个抽象。
答案 0 :(得分:6)
执行此类操作最方便的方法通常是编写自己的类型类。这是一个快速的草图:
import shapeless._
trait T[I, O] extends (I => O)
trait Pipeline[P <: HList] {
type In
type Out
type Values <: HList
}
object Pipeline {
type Aux[P <: HList, In0, Out0, Values0 <: HList] = Pipeline[P] {
type In = In0; type Out = Out0; type Values = Values0
}
def apply[P <: HList](
implicit pipeline: Pipeline[P]
): Aux[P, pipeline.In, pipeline.Out, pipeline.Values] = pipeline
implicit def onePipeline[I, O]: Aux[T[I, O] :: HNil, I, O, I :: O :: HNil] =
new Pipeline[T[I, O] :: HNil] {
type In = I
type Out = O
type Values = I :: O :: HNil
}
implicit def longerPipeline[I, O, P <: HList, Out0, Values0 <: HList](
implicit pipeline: Aux[P, O, Out0, Values0]
): Aux[T[I, O] :: P, I, Out0, I :: Values0] =
new Pipeline[T[I, O] :: P] {
type In = I
type Out = Out0
type Values = I :: Values0
}
}
然后(为了清晰起见重新格式化):
scala> Pipeline[T[String, Int] :: T[Int, Char] :: HNil]
res5: Pipeline[T[String, Int] :: T[Int, Char] :: HNil] {
type In = String
type Out = Char
type Values = String :: Int :: Char :: HNil
} = Pipeline$$anon$2@38fd077c
scala> Pipeline[T[String, Int] :: T[Char, Char] :: HNil]
<console>:19: error: could not find implicit value for parameter
pipeline: Pipeline[[T[String, Int] :: T[Char, Char] :: HNil]
Pipeline[T[String, Int] :: T[Char, Char] :: HNil]
^
无效管道无法编译,对于有效管道,我们可以正确推断出端点和中间值。