Mapped types in Scala

问题:

Is there a way to derive a type from an existing one in Scala?

For example, for case class Person(name: String, age: Int) I'd like to get a Product/Tuple of (Option[String], Option[Int]), i.e. a type mapped from an existing one.

There's a feature in Typescript (mapped types) that allows this relatively easily, which is how I started thinking down this path. But I'm not sure how something like this would be done in Scala.

I feel like the solution involves using shapeless in some way but I'm not sure how to get there.

回答1:

With Shapeless you can define type class

import shapeless.ops.{hlist, product, tuple}
import shapeless.poly.~>
import shapeless.{Generic, HList, Id, the}

trait Partial[A] {
  type Out
}

object Partial {
  type Aux[A, Out0] = Partial[A] { type Out = Out0 }

  object optionPoly extends (Id ~> Option) {
    override def apply[T](t: T): Option[T] = null
  }

//    implicit def mkPartial[A, L <: HList, L1 <: HList](implicit
//      generic: Generic.Aux[A, L],
//      mapper: hlist.Mapper.Aux[optionPoly.type, L, L1],
//      tupler: hlist.Tupler[L1]): Aux[A, tupler.Out] = null

  implicit def mkPartial[A, T](implicit
    toTuple: product.ToTuple.Aux[A, T],
    mapper: tuple.Mapper[T, optionPoly.type],
    ): Aux[A, mapper.Out] = null
}

and use it (the is improved version of implicitly)

case class Person(name: String, age: Int)

// val pp = the[Partial[Person]]
// type PersonPartial = pp.Out

type PersonPartial = the.`Partial[Person]`.Out

implicitly[PersonPartial =:= (Option[String], Option[Int])]

回答2:

I'd suggest parameterize the type as follows:

case class Person[F[_]](name: F[String], age: F[Int])

And then you can derive types you want, like

import cats.Id

type IdPerson = Person[Id]
type OptPerson = Person[Option]

Where cats.Id is simply defined as type Id[A] = A. It is straightforward to write your own, but I suggest using cats' one since it comes with useful typeclass instances.