类型安全：方法..属于原始类型。参考泛型应参数化 [英] Type safety: The method.. belongs to the raw type.. refs to generics should be parameterized

问题描述

Eclipse给了我一个警告（在标题中），在一个工作项目中只使用以下代码，但没有任何东西，但是一个虚拟类和一个主要方法：

Eclipse gives me the warning (in the title) using just the following code in a working project with nothing in it but a dummy class and a main method:

List a = new ArrayList();
List<Integer> b = new ArrayList<Integer>();
int x = 19;
a.add(x);

最后一行触发警告。我不知道我在这里做错什么，还是真的，我甚至在做什么。我是Java学生，遵循可疑的教程，我正在尝试了解泛型。假设变量x没有说明类型的安全性。我尝试将其转换为对象类型（我认为没有，因为它已经是），它没有工作。

The last line triggers the warning. I'm not sure what I am doing wrong here, or really, what I am even doing. I am a Java student following a dubious tutorial and I am attempting to understand generics. Supposedly, variable x doesn't illustrate type safety. I tried casting it to the Object type (which I think does nothing because it already is) and it didn't work.

我如何解决这个警告？ p>

How might I resolve this warning?

推荐答案

您已声明列出没有type参数的。这就是为什么eclipse正在抱怨类型安全性，因为你可以添加任何类型的对象到该列表。

You have declared List a without the type parameter. This is why eclipse is complaining about type safety, as you could add objects of any type to that list.

如果你看看 ArrayList api ，看看类声明，你看到它被声明为这个：

If you look at the ArrayList api and take a look at the class declaration, you see it is declared like this:

public class ArrayList<E>

用任何你想要的课程代替E。

Substitute E with any class you wish.

With List< Integer> b 你已经明确告诉编译器，列表是仅保存 Integer 对象的实例，编译器可以验证这一点，从而给出类型安全。

With List<Integer> b you have explicitly told the compiler that the list is to hold instances of Integer objects only, and the compiler can verify this, thus giving you type safety.

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