Type transformation with shapeless

Question

I have a class similar to this:

class MyClass[T <: HList] {
  val x: ???
}

My problem is the type of the x val. What I'd like is to have it be an HList with each type U of the T HList replaced by Option[U]. I.e. if I specify:

new MyClass[Int :: String :: HNil]

I would like x to have a type of Option[Int] :: Option[String] :: HNil

Is this even possible? How to do it?

Answer 1

You'll need a Mapped instance that witnesses that T and the type of x have this relationship:

import shapeless._, ops.hlist.Mapped

abstract class MyClass[T <: HList, OT <: HList](implicit
  mapped: Mapped.Aux[T, Option, OT]
) {
  val x: OT
}

Unfortunately this is kind of inconvenient to instantiate:

new MyClass[Int :: String :: HNil, Option[Int] :: Option[String] :: HNil] {
  val x = Some(0) :: Some("") :: HNil
}

There are ways around this, but they require some additional changes. For example, you could allow both type parameters to be inferred:

import shapeless._, ops.hlist.Comapped

class MyClass[T <: HList, OT <: HList](val x: OT)(implicit
  mapped: Comapped.Aux[OT, Option, T]
)

And then:

new MyClass(Option(0) :: Option("") :: HNil)

Or you can use something closer to your original class with a custom constructor in the companion object:

import shapeless._, ops.hlist.Mapped

abstract class MyClass[T <: HList] {
  type OT <: HList
  def mapped: Mapped.Aux[T, Option, OT]
  val x: OT
}

object MyClass {
  class PartiallyApplied[T <: HList] {
    def apply[OT0 <: HList](x0: OT0)(implicit
      mapped0: Mapped.Aux[T, Option, OT0]
    ): MyClass[T] =
      new MyClass[T] {
        type OT = OT0
        val mapped: Mapped.Aux[T, Option, OT] = mapped0
        val x: OT = x0
      }
  }

  def apply[T <: HList]: PartiallyApplied[T] = new PartiallyApplied[T]
}

And then:

MyClass[Int :: String :: HNil](Option(0) :: Option("") :: HNil)

Which of these approaches is more appropriate will depend on how you're using the class.