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mapping over HList inside a function

The following code seems obvious enough to compile and run

case class Pair(a: String, b: Int)

val pairGen = Generic[Pair]

object size extends Poly1 {
  implicit def caseInt = at[Int](x => 1)
  implicit def caseString = at[String](_.length)
}

def funrun(p: Pair) = { 
  val hp: HList = pairGen.to(p)
  hp.map(size)
}

but the compiler says "could not find implicit value for parameter mapper". In my use case, I want to map over an HList to get and HList of String(s) and then convert the HList of String(s) into Scala List[String]. Any ideas?

like image 987
arapmv Avatar asked Sep 08 '15 00:09

arapmv


1 Answers

First we can create a Poly1 similar to size which we can use to map an HList to an HList of Strings.

object strings extends Poly1 {
  implicit def caseInt = at[Int](_.toString)
  implicit def caseString = at[String](identity)
}

You were already using Generic[Pair] to turn a Pair into an HList, but you couldn't map over your hp because there is no evidence in your funrun that you can map over it. We can solve this by using implicit parameters.

def funRun[L <: HList, M <: HList](
  p: Pair
)(implicit
  gen: Generic.Aux[Pair, L],
  mapper: Mapper.Aux[strings.type, L, M]
) = gen.to(p).map(strings)
  • Our first implicit parameter gen can turn a Pair into an HList of type L.
  • Our second implicit parameter mapper can use our strings polymorphic function to map an HList of type L to an HList of type M.

We can now use funRun to turn a Pair into an HList of Strings :

scala> funRun(Pair("abc", 12))
res1: shapeless.::[String,shapeless.::[String,shapeless.HNil]] = abc :: 12 :: HNil

But you wanted to return a List[String]. To turn our HList M (the result of mapping to String) to a List[String] we need a ToTraversable, so we add a third implicit parameter :

import shapeless._, ops.hlist._

def pairToStrings[L <: HList, M <: HList](
  p: Pair
)(implicit
  gen: Generic.Aux[Pair, L],
  mapper: Mapper.Aux[strings.type, L, M],
  trav: ToTraversable.Aux[M,List,String]
): List[String] = gen.to(p).map(strings).toList

Which we can use as :

scala> pairToStrings(Pair("abc", 12))
res2: List[String] = List(abc, 12)
like image 200
Peter Neyens Avatar answered Sep 22 '22 11:09

Peter Neyens