[英]Scala lower type bounds and covariance
Im reading this page http://www.scala-lang.org/node/137 , I understand what covariance is and lower bounds as well, but what it's not clear is this line: 我正在阅读这个页面http://www.scala-lang.org/node/137 ,我也明白协方差是什么和下限,但不清楚的是这一行:
Unfortunately, this program does not compile, because a covariance annotation is only possible if the type variable is used only in covariant positions.
不幸的是,这个程序不能编译,因为只有在变量位置使用类型变量时才能进行协方差注释。 Since type variable T appears as a parameter type of method prepend, this rule is broken.
由于类型变量T作为方法前置的参数类型出现,因此该规则被破坏。
why elem
has to be an instance of a supertype of T
, if ListNode
is already covariant why elem
cannot be prepended to the current list. 为什么
elem
必须是超类型T
的实例,如果ListNode
已经是协变的,为什么elem
不能被添加到当前列表中。
class Super {override def toString = "Super"}
class Sub extends Super {override def toString = "Sub"; def subMethod {} }
val sup = new Super
val sub = new Sub
Imagine the following were allowed: 想象一下,允许以下内容:
// invalid code
class Foo[+T] {
def bar(x: T) = println(x)
}
Since Foo
is covariant on T
, this is valid (a simple upcast, since a Foo[Sub]
is a Foo[Super]
): 因为
Foo
在T
上是协变的,所以这是有效的(简单的上传,因为Foo[Sub]
是Foo[Super]
):
val foo : Foo[Super] = new Foo[Sub] {
override def bar(x: Sub) = x.subMethod
}
Now foo
is, as far as we know, a Foo[Super]
like any other, but its bar
method won't work, because the bar
implementation requires a Sub
: 现在
foo
就像我们所知, Foo[Super]
就像其他任何一样,但它的bar
方法不起作用,因为bar
实现需要一个Sub
:
foo.bar(sup) // would cause error!
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