java - LinkedList subList 返回 List 而不是 LinkedList？

Question

我有一个函数f，它获取 a 的子列表LinkedList并将其传递给g，它也需要 a LinkedList：

public static <T> void g(LinkedList<T> l) {
}

public static <T> void f() {
    LinkedList<T> l = new LinkedList<T>();
    ...
    LinkedList<T> l2 = l.subList(i,j);
    ...
    g(l2);
}

但这不会编译，因为显然LinkedList.subList返回List而不是LinkedList. 所以我必须把它改成这样：

LinkedList<T> l2 = (LinkedList<T>)l.subList(i,j);

为什么？

Answer 1

subList is defined in the List interface and returns a List. And if you check the implementation in LinkedList, which is actually in one of the parent classes (AbstractList), you will see that it does not return a LinkedList but a SubList and your cast will throw an exception.

Unless you use methods specific to LinkedList, the easiest way to fix this would be to change g to:

public static <T> void g(List<T> l) {

Answer 2

if g() doesn't use the method of LinkedList but only the List interface is preferable to define g() as follow:

public static <T> void g( List<T> l ) {
}

and the main becomes:

public static <T> void f() {
    List<T> l = new LinkedList<T>();
    ...
    List<T> l2 = l.subList(i,j);
    ...
    g(l2);
}

For a justification, please consult this post.

Answer 3

该方法在由 实现的接口中sublist声明。接口中方法的签名是：ListLinkedListsublistList

 List<E> subList(int fromIndex, int toIndex)

因此，每当您从LinkedList类中调用此方法时，对象将为LinkedList，但方法的返回类型为List，因此将以 的形式给出List。但是，您可以将变量转换为LinkedList并使用它。

Answer 4

您不能实例化“对象”这样的列表。List 是一个接口，而不是一个具体的类。接口只是一个类可以实现的一组功能；实例化一个接口没有任何意义。

希望能帮助到你。

克莱门西奥·莫拉莱斯·卢卡斯。

Answer 5

Java wants as much polymorphism as you can get, so subList returns List, and you have to cast it to what you want.