Scala View 应用程序谜题

Question

假设我们有以下两个特征:

trait Foo[A] { def howMany(xs: List[A]) = xs.size }
trait Bar

以及从第二个到第一个的隐式转换:

implicit def bar2foo[A](bar: Bar) = new Foo[A] {}

我们创建一个Bar和一个整数列表:

val bar = new Bar {}
val stuff = List(1, 2, 3)

现在我希望以下内容能够发挥作用:

bar howMany stuff

但事实并非如此:

scala> bar howMany stuff
<console>:13: error: type mismatch;
 found   : List[Int]
 required: List[A]
              bar howMany stuff
                          ^

所以我们去the spec ，其中有这样的说法(粗体强调是我的):

Views are applied in three situations.

1. [Isn't relevant here.]

2. In a selection e.m with e of type T, if the selector m does not denote a member of T. In this case, a view v is searched which is applicable to e and whose result contains a member named m. The search proceeds as in the case of implicit parameters, where the implicit scope is the one of T. If such a view is found, the selection e.m is converted to v(e).m.

3. In a selection e.m(args) with e of type T, if the selector m denotes some member(s) of T, but none of these members is applicable to the arguments args. In this case a view v is searched which is applicable to e and whose result contains a method m which is applicable to args. The search proceeds as in the case of implicit parameters, where the implicit scope is the one of T. If such a view is found, the selection e.m is converted to v(e).m(args).

因此我们尝试了以下方法，认为这太荒谬了:

trait Foo[A] { def howMany(xs: List[A]) = xs.size }
trait Bar { def howMany = throw new Exception("I don't do anything!") }

implicit def bar2foo[A](bar: Bar) = new Foo[A] {}

val bar = new Bar {}
val stuff = List(1, 2, 3)

但确实如此(至少在 2.9.2 和 2.10.0-RC2 上):

scala> bar howMany stuff
res0: Int = 3

这会导致一些非常奇怪的行为，例如 this workaround对于 this problem .

我有三个(密切相关的)问题:

1. 是否有一种直接的方法(即不涉及添加具有适当名称的虚假方法的方法)来在上述原始情况下正确应用 View ？
2. 有人可以提供解释此行为的规范吗？
3. 假设这是预期的行为，它有任何意义吗？

我也希望能提供有关此问题之前讨论的任何链接 - 我在 Google 上的运气不太好。

Answer 1

供大家引用，这只能是一个bug。您知道的方式是错误消息:

<console>:13: error: type mismatch;
 found   : List[Int]
 required: List[A]

List[A] 不是真正的类型 - 它是应用于其自己的类型参数的 List。这不是一个可以被要求的类型，因为它不是一个可以被表达的类型。

[编辑 - 现在还为时过早，谁知道我在说什么。忽略上述内容，但您仍然可以点击链接。]

与此相关的票证是 https://issues.scala-lang.org/browse/SI-6472 .