Inferred type of function that zips HLists

问题:

Thanks to https://github.com/milessabin/shapeless/wiki/Feature-overview:-shapeless-2.0.0 I understand how to zip shapeless HLists:

Import some stuff from Shapeless 2.0.0-M1:

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

Create two 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

Zip them:

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

Now try to define a function that does the same thing:

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

The inferred return type is Unit, and indeed applying f to h1 and h2 does just that:

scala> f(h1, h2)

scala> 

Is there a way to define f such that I get ((5,6),(a,b)) back in this case?

Ultimately what I'm trying to do is define a function that zips the two HLists and then maps over them, choosing either _1 or _2 based a coin toss, which would yield another HL.

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

Which works fine in the REPL:

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

But I'm getting tripped up on the above issue when trying to pull this into a function.

Thanks!

回答1:

You can wrap everything up in one method using the Zip (or in this case Zip.Aux) type class:

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

Now assuming you have h1 and h2 defined as in the question, you can write this:

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

And so on. This will work in either 2.0.0-M1 or the latest snapshot, although (as I've noted in a comment above) you may run into confusing issues on the way before this bug was fixed.

回答2:

Given the compiler error and some perusing the tests in hlist.scala, zip is defined this way:

def f[L <: HList, OutT <: HList](l : L)(
  implicit transposer : Transposer.Aux[L, OutT],
           mapper : Mapper[tupled.type, OutT]) = l.transpose.map(tupled)

And the application of my mix can be defined this way:

def g[L <: HList](l : L)(
  implicit mapper: Mapper[mix.type,L]) = l.map(mix)

The composition does what I was looking for:

scala> g(f(h1 :: h2 :: HNil))
res12: shapeless.::[Int,shapeless.::[String,shapeless.HNil]] = 5 :: b :: HNil

scala> g(f(h1 :: h2 :: HNil))
res13: shapeless.::[Int,shapeless.::[String,shapeless.HNil]] = 6 :: a :: HNil

scala> g(f(h1 :: h2 :: HNil))
res14: shapeless.::[Int,shapeless.::[String,shapeless.HNil]] = 6 :: b :: HNil