问题描述
private List gridModel; public List getGridModel() { return gridModel; }
Eclipse 显示警告:
Eclipse shows a warning:
List 是原始类型.对泛型 List 的引用应该被参数化.
List is a raw type. References to generic type List should be parameterized.
将代码更改为以下将删除警告
Changing the code to below will remove the warning
private List<?> gridModel; public List<?> getGridModel() { return gridModel; }
但是,上面的代码在 SonarQube 中显示了一个主要的陷阱错误:
However the above code shows a Major pitfall error in SonarQube which says:
删除通用通配符类型的使用.返回参数中不应使用通用通配符类型
Remove usage of generic wildcard type. Generic wildcard types should not be used in return parameters
那么我该如何解决这个警告呢?
我在这里看到 类似问题,但可以找不到解决办法.
So how can I fix this warning?
I see a similar question here but could not find the solution .
使用 Class 没有移除声纳警告.
Using Class<? extends Object> did not remove Sonar warning.
推荐答案
那么我该如何解决这个警告呢?
So how can I fix this warning ?
你可以为你的类使用类型参数:
You can use a type parameter for your class :
public class GridModelHolder<T> { private List<T> gridModel; public List<T> getGridModel() { return gridModel; } }
然后客户端代码可以决定 List GridModelHolder 的类型:
The client code can then decide what type of List GridModelHolder holds :
GridModelHolder<字符串>gridModelHolder = new GridModelHolder
但是,如果您坚持使用原始类型,您可以取消警告或仅使用对象列表(不推荐使用这两种方法)
However, if you insist on using raw types, you can either suppress the warnings or simply have a List of objects (Neither of these are recommended)
@SuppressWarnings("unchecked") public class GridModelHolder { private List gridModel; public List getGridModel() { return gridModel; } }
或
public class GridModelHolder { private List<Object> gridModel; public List<Object> getGridModel() { return gridModel; } }