从模式匹配返回路径相关类型

Question

鉴于异构类型:

trait Request {
  type Result
}

trait IntRequest extends Request {
  type Result = Int
}

如何让Scala编译器满意地根据模式匹配返回路径依赖类型:

def test(in: Request): in.Result = in match {
  case i: IntRequest => 1234
  case _ => sys.error(s"Unsupported request $in")
}

错误:

<console>:53: error: type mismatch;
 found   : Int(1234)
 required: in.Result
         case i: IntRequest => 1234
                               ^

Answer 1

以下作品:

trait Request {
  type Result
}

final class IntRequest extends Request {
  type Result = Int
}

trait Service {
  def handle[Res](in: Request { type Result = Res }): Res
}

trait IntService extends Service {
  def handle[Res](in: Request { type Result = Res }): Res = in match {
    case i: IntRequest => 1234
    case _ => sys.error(s"Unsupported request $in")
  }
}

trait Test {
  def service: Service

  def test(in: Request): in.Result = service.handle[in.Result](in)
}

如果使用a final class,编译器只会吃它吗？