推断HLists的推断类型

Question

感谢https://github.com/milessabin/shapeless/wiki/Feature-overview:-shapeless-2.0.0我理解如何压缩无形的HLists:

从Shapeless 2.0.0-M1导入一些东西:

import shapeless._
import shapeless.ops.hlist._
import syntax.std.tuple._
import Zipper._

创建两个HLists:

scala> val h1 = 5 :: "a" :: HNil
h1: shapeless.::[Int,shapeless.::[String,shapeless.HNil]] = 5 :: a :: HNil

scala> val h2 = 6 :: "b" :: HNil
h2: shapeless.::[Int,shapeless.::[String,shapeless.HNil]] = 6 :: b :: HNil

拉链他们:

scala> (h1, h2).zip
res52: ((Int, Int), (String, String)) = ((5,6),(a,b))

现在尝试定义一个执行相同操作的函数:

scala> def f[HL <: HList](h1: HL, h2: HL) = (h1, h2).zip
f: [HL <: shapeless.HList](h1: HL, h2: HL)Unit

推断的返回类型是Unit,实际上将f应用于h1和h2就是这样:

scala> f(h1, h2)

scala> 

有没有办法定义f,以便在这种情况下得到((5,6),(a,b))？

最终我要做的是定义一个拉伸两个HList然后映射它们的函数,选择基于抛硬币的_1或_2,这将产生另一个HL.

object mix extends Poly1 {
  implicit def caseTuple[T] = at[(T, T)](t =>
    if (util.Random.nextBoolean) t._2 else t._1)
}

哪个在REPL中工作正常:

scala> (h1, h2).zip.map(mix)
res2: (Int, String) = (5,b)

但是当我试图把它拉成一个函数时,我正在惹恼上面的问题.

谢谢!

Answer 1

您可以使用Zip(或在本例中Zip.Aux)类型类在一个方法中包装所有内容:

import shapeless._, shapeless.ops.hlist._

object mix extends Poly1 {
  implicit def caseTuple[T] = at[(T, T)](t =>
    if (util.Random.nextBoolean) t._2 else t._1)
}

def zipAndMix[L <: HList, Z <: HList](h1: L, h2: L)(implicit
  zipper: Zip.Aux[L :: L :: HNil, Z],
  mapper: Mapper[mix.type, Z]
) = (h1 zip h2) map mix

现在假设您已经h1并h2在问题中定义,您可以这样写:

scala> zipAndMix(h1, h2)
res0: shapeless.::[Int,shapeless.::[String,shapeless.HNil]] = 5 :: b :: HNil

scala> zipAndMix(h1, h2)
res1: shapeless.::[Int,shapeless.::[String,shapeless.HNil]] = 6 :: a :: HNil

scala> zipAndMix(h1, h2)
res2: shapeless.::[Int,shapeless.::[String,shapeless.HNil]] = 5 :: a :: HNil

等等.这将在2.0.0-M1或最新快照中工作,尽管(正如我在上面的评论中所述),在修复此错误之前,您可能会遇到令人困惑的问题.