Java 未绑定通配符泛型

发布于 2025-01-03 03:42:58 字数 214 浏览 0 评论 0原文

与完全跳过通配符类型泛型相比，在 Bar 类中使用通配符类型泛型有什么优势吗？

public class Foo<T> {}

public interface Bar {
    public void addFoo(Foo<?> foo);
    public Foo<?> getFoo(String name);
}

Are there any advantages of using wildcard-type generics in the Bar class over completely skipping them?

public class Foo<T> {}

public interface Bar {
    public void addFoo(Foo<?> foo);
    public Foo<?> getFoo(String name);
}

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帅气称霸 2025-01-10 03:42:58

有多种优点。

它们不会像使用原始类型那样产生编译器警告
它们提供了更多的类型安全性。例如，考虑 Foo 是否为 List。如果您使用 List 而不是 List，您可以这样做：
```
myBar.getFoo("数字").add("某个字符串");
```
即使列表仅应该包含数字。如果您返回一个 List，那么您将无法向其中添加任何内容（null 除外），因为列表未知。
它们记录了与原始类型完全不同的内容，即 Foo 是在某些未知但特定的类型上键入的。

There are multiple advantages.

They won't produce compiler warnings like using the raw type would
They give more type safety. For example, consider if Foo was List instead. If you used List instead of List<?>, you could do this:
```
myBar.getFoo("numbers").add("some string");
```
even if the list was only supposed to contain Numbers. If you returned a List<?>, then you would not be able to add anything to it (except null) since the type of list is not known.
They document something completely different than a raw type, namely that Foo is typed on some unknown but specific type.