Scala 示例 - 具有上下文边界错误的特征类型参数？

发布于 2024-12-22 14:34:23 字数 391 浏览 2 评论 0原文

阅读《Scala by Example》一书，Martin 在第 54 页以及

trait Set[A <: Ordered[A]] {
  def incl(x: A): Set[A]
  def contains(x: A): Boolean
}

第

trait Set[A <% Ordered[A]] ...

55 页上解释了类型界限时，有一个示例。他还说 <:/<% 是为了证明特征集所需的唯一更改类型限制的可能性。

然而，当我用自己的代码重复该示例时，IDE 抱怨特征可能没有视图边界，只有类型边界。将特征关键字更改为抽象类或将视图绑定更改为类型绑定会有所帮助。这是书上的错误吗？

原文

Reading the Scala by Example book and there is this example when Martin explains type bounds on page 54:

trait Set[A <: Ordered[A]] {
  def incl(x: A): Set[A]
  def contains(x: A): Boolean
}

and

trait Set[A <% Ordered[A]] ...

further on page 55. He also says that the <:/<% is the only change required for trait Set in order to demonstrate the type bounding possibilities.

However, when I repeat the example with my own code, the IDE complains that traits may NOT have view bounds, only type bounds. Changing the trait keyword to abstract class or changing the view bound to type bound helps. Is this a mistake in the book?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

捂风挽笑 2024-12-29 14:34:23

让我们使用名为 REPL 的强大工具来了解正在发生的事情：

scala>  trait Example[A<:Ordered[A]] { def contains(x:A):Boolean }
defined trait Example

scala>  class Example2[A<%Ordered[A]]( val a:A) { def isLower(otherA:A):Boolean = a< otherA }
defined class Example2

scala>  :javap Example
Compiled from "<console>"
public interface Example{
    public abstract boolean contains(scala.math.Ordered);
}


scala>  :javap Example2
Compiled from "<console>"
public class Example2 extends java.lang.Object implements scala.ScalaObject{
    public java.lang.Object a();
    public boolean isLower(java.lang.Object);
    public Example2(java.lang.Object, scala.Function1);
}

如您所见，视图绑定成为 Example2 构造函数的第二个参数。由于特征没有构造函数，因此显然不可能提供视图绑定。

为什么这在以前的版本中是可能的，对我来说是一个谜（也许在特征内创建了一个额外的 Function1 val 并由编译器填充？）

关于您关于 Scala 演变的问题，它是成熟且强大的。您可以预期主要版本（2.8、2.9、2.10）之间的变化，但我不认为 scala 对此还不够成熟。然而，总是有改进的空间

Let's use our powerful tool called the REPL to understand what is going on:

scala>  trait Example[A<:Ordered[A]] { def contains(x:A):Boolean }
defined trait Example

scala>  class Example2[A<%Ordered[A]]( val a:A) { def isLower(otherA:A):Boolean = a< otherA }
defined class Example2

scala>  :javap Example
Compiled from "<console>"
public interface Example{
    public abstract boolean contains(scala.math.Ordered);
}


scala>  :javap Example2
Compiled from "<console>"
public class Example2 extends java.lang.Object implements scala.ScalaObject{
    public java.lang.Object a();
    public boolean isLower(java.lang.Object);
    public Example2(java.lang.Object, scala.Function1);
}

As you can see , the view bound becomes the second argument of the Example2 constructor. Since a trait does not have a constructor, clearly it is not possible to provide a view bound.

Why this has been possible in previous releases is to me a mistery (maybe an additional Function1 val was created inside the trait and filled by the compiler?)

Concerning your question about Scala evolution, it is mature and powerful. You can expect changes between major releases (2.8, 2.9, 2.10) but I would not consider scala not mature enough for this. There is, however, always space for improvement

回复收藏 0 原文

~没有更多了~