java - 如何将对象添加到未参数化的 Java 列表？

Question

我有一个 Java 对象 OldFashioned扩展 Java 1.4 List :

[Java] 
class OldFashioned extends List { ... }

即OldFashioned不接受任何类型参数。我需要添加 SomeObject给它。在 Java 中没有问题，因为它处理 List 1.4 作为 List<Object> 1.5 并允许添加 Object 的任何子类到集合。但是斯卡拉没有。所以下一个代码不起作用:

[Scala]
val oldFashioned = new OldFashioned()
oldFashioned.add(new SomeObject)       // found: SomeObject; required: E

也就是说，Scala 编译器 需要将类型参数传递给 OldFashioned这实际上并没有带他们:

[Scala]
var oldFashioned: OldFashioned[SomeObject] = null    // OldFashioned does not take type parameters

我怎样才能克服它并添加 SomeObject至 OldFashined ？

Answer 1

好吧，我不敢相信我的 earlier question在这里找到一个用法(你应该去投票 agilesteel 的 answer )

def add(oldFashioned: OldFashioned, any: Any): Boolean = oldFashioned match {
  case l: java.util.List[a] => l.add(any.asInstanceOf[a])
}

val oldFashioned = new OldFashioned
// oldFashioned: OldFashioned = []

add(oldFashioned, "test")
// res0: Boolean = true
add(oldFashioned, 1)
// res1: Boolean = true
add(oldFashioned, new Object)
// res2: Boolean = true

oldFashioned
// res3: OldFashioned = [test, 1, java.lang.Object@1db8f8e]

编辑:我猜只要我要投:

oldFashioned.asInstanceOf[java.util.List[SomeObject]].add(new SomeObject)