如何初始化List< String> Java中的对象？

问题描述：

我无法按以下代码初始化List：

I can not initialize a List as in the following code:

List<String> supplierNames = new List<String>();
supplierNames.add("sup1");
supplierNames.add("sup2");
supplierNames.add("sup3");
System.out.println(supplierNames.get(1));

我遇到以下错误：

无法实例化类型列表< String>

我如何实例化列表< String> ？

答

如果你查看 API >列表你会注意到它：

If you check the API for List you'll notice it says:

Interface List<E>

作为界面表示无法实例化（没有新的List（）是可能的。）

Being an interface means it cannot be instantiated (no new List() is possible).

如果你检查那个链接，你会发现一些 class es实现列表：

If you check that link, you'll find some classes that implement List:

所有已知的实施类：

All Known Implementing Classes:

AbstractList ， AbstractSequentialList ， ArrayList ， AttributeList ， CopyOnWriteArrayList ， LinkedList ， RoleList ， RoleUnresolvedList ， Stack ， Vector

AbstractList, AbstractSequentialList, ArrayList, AttributeList, CopyOnWriteArrayList, LinkedList, RoleList, RoleUnresolvedList, Stack, Vector

可以实例化这些。使用他们的链接了解更多关于他们的信息，IE：知道哪个更适合您的需求。

Those can be instantiated. Use their links to know more about them, I.E: to know which fits better your needs.

3个最常用的可能是：

The 3 most commonly used ones probably are:

 List<String> supplierNames1 = new ArrayList<String>();
 List<String> supplierNames2 = new LinkedList<String>();
 List<String> supplierNames3 = new Vector<String>();

奖金：

你也可以使用 Arrays class 以更简单的方式使用值实例化它，如下所示：

Bonus:
You can also instantiate it with values, in an easier way, using the Arrays class, as follows:

List<String> supplierNames = Arrays.asList("sup1", "sup2", "sup3");
System.out.println(supplierNames.get(1));

但请注意，您不能在该列表中添加更多元素，因为它是固定大小。

But note you are not allowed to add more elements to that list, as it's fixed-size.

如何初始化List< String> Java中的对象？

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