Scala更高的kinded类型方差

Question

我正在用更高级的类型浸泡我的脚趾,探索一个非常基本的Scala示例:

trait Mappable[F[_]] {
  def map[A, B](fa: F[A])(f: A => B): F[B]
}

object Mappable {
  implicit object MappableOption extends Mappable[Option] {
    def map[A, B](fa: Option[A])(f: A => B): Option[B] = fa.map(f)
  }
  implicit object MappableSeq extends Mappable[Seq] {
    def map[A, B](fa: Seq[A])(f: A => B): Seq[B] = fa.map(f)
  }
}

def bananaTuple[F[_], T](f: F[T])(implicit F: Mappable[F]): F[(String, T)] =
  F.map(f)(("banana", _))

这有效:

bananaTuple(Option(42)) // Some((banana,42))
bananaTuple(Seq(42))    // List((banana,42))

但这不编译:

bananaTuple(Some(42))
bananaTuple(List(42))

我得到的编译错误:

could not find implicit value for parameter F: ch.netzwerg.hkt.HigherKindedTypes.Mappable[Some] bananaTuple(Some(42))

not enough arguments for method bananaTuple: (implicit F: ch.netzwerg.hkt.HigherKindedTypes.Mappable[Some])Some[(String, Int)]. Unspecified value parameter F. bananaTuple(Some(42))

如何在游戏中带来变化？

Answer 1

我们可以通过更多的参数多态性来完成这项工作：

object MappableExample {
  trait Mappable[F[_]] {
    type Res[_]
    def map[A, B](f: A => B)(c: F[A]): Res[B]
  }

  implicit def seqMappable[C[X] <: Seq[X]] = new Mappable[C] {
    type Res[X] = Seq[X]
    override def map[A, B](f:A => B)(c: C[A]): Seq[B] = c.map(f)
  }

  implicit def optionMappable[C[X] <: Option[X]]: Mappable[C] = new Mappable[C] {
    type Res[X] = Option[X]
    override def map[A, B](f: A => B)(c: C[A]): Option[B] = c.map(f)
  }

  def map[A, B, C[_]](xs: C[A])(f: A => B)(implicit mappable: Mappable[C]): mappable.Res[B] = {
    mappable.map(f)(xs)
  }

  def main(args: Array[String]): Unit = {
    println(map(List(1,2,3))(("banana", _)))
    println(map(Some(1))(("banana", _)))
  }
}

产量：

List((banana,1), (banana,2), (banana,3))
Some((banana,1))

编译器现在推断出SomeasMappable[Some]#Res[Int]和 ，Mappable[List]#Res[Int]这非常难看。人们期望编译器实际上能够推断出正确的类型，而不需要特征上的任何协变/逆变Mappable，但我们不能这样做，因为我们在不变的位置使用它。