scala - case语句中的类型推断失败

Question

据我所知，这是最小的案例。

import akka.actor.ActorDSL.Act
trait Foo[A]

object Foo {
  case class Str(str: String) extends Foo[String]
  case object Actor extends Foo[Act]

  def foo[A](f: Foo[A]): A = f match {
    case Str(string) => string
    case Actor => new Act {
      become { case e => println(e) }
    }
  }
}

通过此错误

Foo.scala:12: type mismatch;
[error]  found   : akka.actor.ActorDSL.Act
[error]  required: A
[error]     case Actor => new Act {
[error]                   ^

Str大小写只是用来说明类型推断在正常用例中有效，但是Scala某种程度上“忘记”了A在检查case语句的返回类型时应该是特定类型。

请注意，如果将new Act中的部分函数提取为单独的函数，它将再次正确编译。

import akka.actor.ActorDSL.Act
trait Foo[A]

object Foo {
  case class Str(str: String) extends Foo[String]
  case object Actor extends Foo[Act]

  val partial: PartialFunction[Any, Unit] = {
    case e => println(e)
  }

  def foo[A](f: Foo[A]): A = f match {
    case Str(string) => string
    case Actor => new Act {
      become(partial)
    }
  }
}

这是编译器错误吗？

Answer 1

您的导入必须有问题，因为否则示例将正确编译:

trait Foo[A]
trait Act

object Foo {
  case class Str(str: String) extends Foo[String]
  case object Actor extends Foo[Act]

  def foo[A](f: Foo[A]): A = f match {
    case Str(string) => string
    case Actor => new Act {}
  }
}

在Scala 2.11.7中测试。