scala – 在通配符上键入Boundary“Stickyness”

这是一个回顾：

I just realized that the Scala compiler does not seem to “carry over” type boundaries on wildcards:

scala> class Foo { def foo = 42 }
defined class Foo
scala> class Bar[A <: Foo](val a: A)
defined class Bar
scala> def bar(x: Bar[_]) = x.a.foo
:7: error: value foo is not a member of Any

I would expect that the parameter of method bar is still upper-bound by Foo,even though its exact type parameter is unimportant in the method.
Is there a specific reason for this behavior?

然后讨论进入Spec解释争议:)

说明？

最后海报提出了这样的解释：

However,if the compiler did this for Bar[_],for consistency reasons it would also have to do it for Bar[A]. The latter however would have some strange consequences.
def bar[A](x: Bar[A],y: Bob[A])
for example would suddenly have to carry an implicit type bound for Bob. If Bob had its own type bound things would be really messy.

我不明白,因为

def bar[A](x: Bar[A])

因为Bar类型参数是有界的,所以不会单独编译.

无论如何,我相信以下将是完全有道理的：

def bar(x: Bar[_],y : Bob[_])

解决方法

作为一种解决方法,他们建议：

def bar(x: Bar[_ <: Foo]) = x.a.foo

除了不干燥之外,它会让事情变得困难：

让我们考虑一棵树

abstract class Tree[T <: Tree[T]] { val subTree : List[T] }

现在,您将如何定义递归遍历树的函数(显然在Tree类之外定义)：

def size( tree : Tree[_] ) = tree.subTree.size + tree.subTree.map(size(_)).sum

显然不会工作,因为subTree将变成List [Any],所以我们需要一个类型参数：

def size[T <: Tree[T]]( tree : T ) = ...

甚至更丑陋：

class OwnATree( val myTree : Tree[_] ){}

应该成为

class OwnATree[T <: Tree[T]]( val myTree : T ){}

等等……

我相信某处出了点问题:)