我正在学习 Shapeless 并实现了一个简单的异构树
我希望能够提取保留类型信息的标签。我很努力，但还没有草图来展示解决方案。任何帮助表示赞赏!
object TreeTest {
import shapeless._
case class HTree[A, F <: HList](label: A, forest: F) {
def withBranch[T, FF <: HList](tree: HTree[T, FF]): HTree[A, HTree[T, FF] :: F] = new HTree(label, tree :: forest)
}
object HTree {
def apply[A](label: A) = new HTree[A, HNil](label, HNil)
}
val t1 = HTree(1)
val t2 = HTree("1")
val t3 = t2.withBranch(HTree(2.0))
val t4: HTree[Int, HTree[String, HTree[Double, HNil] :: HNil] :: HNil] = t1.withBranch(t3)
def labels[A, F <: HList](t: HTree[A, F]) = ???
val flattened: Int :: String :: Double :: HNil = labels(t4)
}
最佳答案
我能够创建一个递归多态函数来提取您的 HTree 的标签:
import shapeless._, ops.hlist._
object getLabels extends Poly1 {
implicit def caseHTree[A, F <: HList, M <: HList](implicit
fm: FlatMapper.Aux[getLabels.type, F, M],
prepend: Prepend[A :: HNil, M]
) : Case.Aux[HTree[A, F], prepend.Out] =
at[HTree[A, F]](tree => prepend(tree.label :: HNil, fm(tree.forest)))
}
它将当前 HTree 的标签添加到它递归获取的分支的标签(在 forest 上进行平面映射)。
t4 的结果:
getLabels(t4)
// Int :: String :: Double :: HNil = 1 :: 1 :: 2.0 :: HNil
关于scala - 异构树的类型级展平，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/33426077/