如何初始化 List<String>Java中的对象? [英] How to initialize List&lt;String&gt; object in Java?

问题描述

我不能像下面的代码那样初始化一个列表:

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List

如何实例化List?

推荐答案

如果您查看 API for List 你会注意到它说:

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

Interface List<E>

作为 interface 意味着它不能被实例化(没有 new 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，你会发现一些实现List的class:

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

所有已知的实现类:

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

其中一些可以实例化(未定义为抽象类的那些).使用他们的链接了解更多关于他们的信息，即:了解哪个更适合您的需求.

Some of those can be instantiated (the ones that are not defined as abstract class). Use their links to know more about them, I.E: to know which fits better your needs.

最常用的 3 个大概是:

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>();

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奖励:
您还可以使用 Arrays class 以更简单的方式用值实例化它，如下所示:

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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，您不能向该列表添加更多元素，因为它是fixed-size.

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

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