Question

我正在尝试使用案例对象＆＃39;类型作为抽象类型。我很惊讶地看到下面的（类似）代码编译：

sealed abstract class Bar

case object BarOne extends Bar

case object BarTwo extends Bar

sealed abstract class Foo {
  type A <: Bar

  def f: A
}

object Foo {
  object FooOne extends Foo {
    type A = BarOne.type
    val f = BarTwo
  }

  object FooTwo extends Foo {
    type A = BarTwo.type
    val f = BarOne
  }
}

在我的实例中，Foo被参数化并用作案例类。所以我不能只让A成为一个类型参数。

当f = BarTwo设置为A时，BarOne.type如何编译？

如果A中的f: A被解释为A <: Bar，为什么会如此？

有没有办法为A的每个对象实例具体设置Foo？

我正在使用Scala 2.11.8。

用val attributeType = ...中的def attributeType = ...替换FooOne时，

更新 FooTwo编译失败（正如预期的那样）。

Answer 1

我不知道这里发生了什么，但我确实让问题更加孤立。此外，它适用于Foo子类和对象。我已经在scalac 2.11.8上证实了这个编译：

object BarOne
object BarTwo

abstract class Foo[A] {
  def attributeType: A
}

object FooContainer {
  class FooOne extends Foo[BarOne.type] {
    val attributeType = BarTwo
  }

  object FooTwo extends Foo[BarOne.type] {
    val attributeType = BarOne
  }
}

Answer 2

有人建议您升级到Scala的现代版本吗？（笑话。）

有关覆盖的错误为该类型提供了一个很好的路径。

$ scala
Welcome to Scala 2.12.0-M5 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_92).
Type in expressions for evaluation. Or try :help.

scala> :pa
// Entering paste mode (ctrl-D to finish)

sealed abstract class Bar

case object BarOne extends Bar

case object BarTwo extends Bar

sealed abstract class Foo {
  type A <: Bar

  def f: A
}

object Foo {
  object FooOne extends Foo {
    type A = BarOne.type
    val f = BarTwo
  }

  object FooTwo extends Foo {
    type A = BarTwo.type
    val f = BarOne
  }
}

// Exiting paste mode, now interpreting.

<console>:26: error: overriding method f in class Foo of type => Foo.FooOne.A;
 value f has incompatible type
           val f = BarTwo
               ^
<console>:31: error: overriding method f in class Foo of type => Foo.FooTwo.A;
 value f has incompatible type
           val f = BarOne
               ^

错误是this one，重复的问题是from last November。

这个bug的本质也很聪明：它是由-Yoverride-objects引入的，这不是一个非常有用的选项，但在我的几个S.O.答案，现在问题。

编辑：

$ scala
Welcome to Scala 2.11.8 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_92).
Type in expressions for evaluation. Or try :help.

scala> object X ; object Y
defined object X
defined object Y

scala> class C { def f: X.type = X }
defined class C

scala> class D extends C { override def f: Y.type = Y }
defined class D

scala> :quit
$ scalam
Welcome to Scala 2.12.0-M5 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_92).
Type in expressions for evaluation. Or try :help.

scala> scala> object X ; object Y

// Detected repl transcript. Paste more, or ctrl-D to finish.

defined object X
defined object Y

scala> class C { def f: X.type = X }
defined class C

scala> class D extends C { override def f: Y.type = Y }
defined class D
// Replaying 3 commands from transcript.

scala> object X ; object Y
defined object X
defined object Y

scala> class C { def f: X.type = X }
defined class C

scala> class D extends C { override def f: Y.type = Y }
<console>:13: error: overriding method f in class C of type => X.type;
 method f has incompatible type
       class D extends C { override def f: Y.type = Y }
                                        ^

Answer 3

在f中：A被解释为A＆lt ;: Bar
因为当在抽象类

如何覆盖抽象定义（以下内容无法编译）：

object Foo {

  object FooOne extends Foo {
    type A = BarOne.type
    override val f: A = BarTwo
  }

  object FooTwo extends Foo {
    type A = BarTwo.type
    override val f: A = BarOne
  }

}

如何具体设置类型绑定的抽象类型？

3 个答案: