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[英]How to transform an HList to another HList with foldRight/foldLeft
[英]HList foldLeft with tuple as zero
我试图在带有(HL, Int)
类型的累加器的foldLeft
上 foldLeft ,其中HL
是 HList。 下面的程序不能编译。 但是,如果我切换到更简单的HL
类型累加器(只需将注释行与上面的行切换),它就会编译并且可以工作。
将 HList 包装在元组中会破坏 leftFolder 的隐式解析。 我错过了什么?
package foo.bar
import shapeless.{:+:, ::, CNil, Coproduct, Generic, HList, HNil, Lazy, Poly2}
import shapeless.ops.hlist.{LeftFolder, Reverse}
object StackOverflow extends App {
trait MyTypeclass[T] {
def doSomething(t: T): (T, Int)
}
implicit lazy val stringInstance: MyTypeclass[String] = (t: String) => (t, 0)
implicit val hnilInstance: MyTypeclass[HNil] = (t: HNil) => (t, 0)
implicit def hlistInstance[H, T <: HList](
implicit
head: Lazy[MyTypeclass[H]],
tail: MyTypeclass[T]
): MyTypeclass[H :: T] =
(ht: H :: T) =>
ht match {
case h :: t =>
val (hres, hint) = head.value.doSomething(h)
val (tres, tint) = tail.doSomething(t)
(hres :: tres, hint + tint)
}
implicit val cnilInstance: MyTypeclass[CNil] = (t: CNil) => ???
implicit def coproductInstance[L, R <: Coproduct](
implicit
head: Lazy[MyTypeclass[L]],
tail: MyTypeclass[R]
): MyTypeclass[L :+: R] = (lr: L :+: R) => ???
object leftFolder extends Poly2 {
implicit def caseAtSimple[F, HL <: HList]: Case.Aux[HL, F, F :: HL] =
at {
case (acc, f) => f :: acc
}
implicit def caseAtComplex[F, HL <: HList]: Case.Aux[(HL, Int), F, (F :: HL, Int)] =
at {
case ((acc, i), f) => (f :: acc, i)
}
}
implicit def genericInstance[T, HL <: HList, LL <: HList](
implicit
gen: Generic.Aux[T, HL],
myTypeclass: Lazy[MyTypeclass[HL]],
// folder: LeftFolder.Aux[HL, HNil, leftFolder.type, LL],
folder: LeftFolder.Aux[HL, (HNil, Int), leftFolder.type, (LL, Int)],
reverse: Reverse.Aux[LL, HL]
): MyTypeclass[T] = (t: T) => {
val generic = gen.to(t)
val (transformed, idx) = myTypeclass.value.doSomething(generic)
// val ll = transformed.foldLeft(HNil: HNil)(leftFolder)
val (ll, _) = transformed.foldLeft((HNil: HNil, 0))(leftFolder)
val reversed = reverse(ll)
(gen.from(reversed), idx)
}
def doSomething[T](t: T)(implicit myTypeclass: MyTypeclass[T]): T = myTypeclass.doSomething(t)._1
case class Foo(
str1: String,
str2: String
)
val original = Foo("Hello World!", "Hello there!")
val result = doSomething(original)
println(result == original)
}
您希望隐式在一个步骤中完成太多工作。
尝试再添加一个类型参数Out
implicit def genericInstance[T, HL <: HList, Out, LL <: HList](
implicit
gen: Generic.Aux[T, HL],
myTypeclass: Lazy[MyTypeclass[HL]],
//folder: LeftFolder.Aux[HL, (HNil, Int), leftFolder.type, (LL, Int)],
folder: LeftFolder.Aux[HL, (HNil, Int), leftFolder.type, Out],
ev: Out <:< (LL, Int), // added
reverse: Reverse.Aux[LL, HL]
): MyTypeclass[T] = (t: T) => {
val generic = gen.to(t)
val (transformed, idx) = myTypeclass.value.doSomething(generic)
//val (ll, _) = transformed.foldLeft((HNil: HNil, 0))(leftFolder)
val (ll, _) = ev(transformed.foldLeft((HNil: HNil, 0))(leftFolder))
val reversed = reverse(ll)
(gen.from(reversed), idx)
}
阅读过度约束的隐式:
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