Question

我试图抽象一些可以返回任何类型A，Option[A]或Seq[A]的库API。

到目前为止，我有这样的事情：

  type Const[T] = T

  sealed abstract class Request[F[_], A]

  case class GetOne(id: Int) extends Request[Const, Int]
  case class GetMany() extends Request[Seq, String]

然后当我使用它时：

def get[F[_], A](request: Request[F, A]): F[A] = request match {
  case GetOne(id) => client.getOne[F[A]](id)
  case GetMany() => client.getMany[A]() // error: Seq[A] does not conform to F[A]
}

我理解为什么这不起作用F[_]不是covariant Seq[_]的子类或类似的东西。但是，我仍然不确定如何在能够使用Const[A]的同时解决问题。我绝望了吗？请帮忙。

Answer 1

对于这种类型的多态，你可以使用类型类概念
考虑

trait Client {
  def getOne[X]: X
  def getMany[X]: Seq[X]
}

type Const[T] = T

sealed abstract class Request[F[_], A]

case class GetOne(id: Int) extends Request[Const, Int]
case class GetMany() extends Request[Seq, String]

我们可以定义这样的类型类：

trait HandleRequest[R <: Request[F, A], F[_], A] {
  def apply(request: R, client: Client): F[A]
}

并针对所需案例进行实例化：

implicit object handleGetOne extends HandleRequest[GetOne, Const, Int] {
  def apply(request: GetOne, client: Client): Int = client.getOne
}

implicit object handleGetMany extends HandleRequest[GetMany, Seq, String] {
  def apply(request: GetMany, client: Client): Seq[String] = client.getMany
}

现在您可以按如下方式定义常规功能：

implicit class ClientOps(val client: Client) {
  def get[R <: Request[F, A], F[_], A](request: R)(implicit handle: HandleRequest[R, F, A]): F[A] =
    handle(request, client)
}

如果您想要概括您的请求类型，例如：

case class GetOne[X](id: Int) extends Request[Const, X]
case class GetMany[X]() extends Request[Seq, X]

您可以将实例重新定义为：

implicit def handleGetOne[X] = new HandleRequest[GetOne[X], Const, X] {
  def apply(request: GetOne[X], client: Client): X = client.getOne
}

implicit def handleGetMany[X] = new HandleRequest[GetMany[X], Seq, X] {
  def apply(request: GetMany[X], client: Client): Seq[X] = client.getMany
}

Answer 2

几个月后回到这里，我已经意识到我可以使用path-dependent type来实现相同的结果，同时更简洁，不需要类型类型模式。

type Const[T] = T

sealed trait Request {
  type F[_]
  type A
  type FA = F[A]

  def query(client: Client): Future[FA]
}

case class GetOne(id: Int) extends Request {
  type F[x] = Const[x]
  type A = Int
  def query(client: Client): Future[Int] = client.getOne(id)
}

case class GetMany(id: Int) extends Request {
  type F[x] = Seq[x]
  type A = String
  def query(client: Client): Future[Seq[String]] = client.getMany(id)
}

然后我们可以在没有类型参数爆炸的情况下调用它：

def get[R <: Request](request: R): request.FA = request.query(client)

Scala更高级的类型和协变

2 个答案: