Scala视图应用程序难题

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

并从第二个到第一个隐含的转换：

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)

现在我期待以下工作：

bar howMany stuff

但它没有：

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,这就是说(粗体重点是我的)：

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).

所以我们尝试以下,认为工作太荒谬了,

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(1,3)

但是(至少在2.9.2和2.10.0-RC2上)：

scala> bar howMany stuff
res0: Int = 3

这导致了一些非常奇怪的行为,例如this workaround的this problem.

我有三个(密切相关)的问题：

>有没有一个直截了当的方式(即不涉及添加具有适当名称的假方法)的方式,以便在上述原始情况下正确应用视图？
有人可以提供说明这个行为的规范吗？
假设这是预期的行为,它有什么意义吗？

我也很感激与以前讨论这个问题的任何链接 – 我对Google没有太多的好运气.