java - 我可以在通用类型上获取 Clazz.class 吗？

Question

我正在尝试调用 EasyMock.isA(Class<T>)在 List<MyType> 上.有没有办法在没有警告的情况下做到这一点？

我试过以下方法:

isA(List<MyType>.class);  // doesn't compile
isA(List.<MyType>class);  // syntax error on tokens (MyType), misplaced construct
isA(List.class);          // This gives a warning: Type safety: The expression of type List needs unchecked conversion to conform to List<MyType>

编辑:

Jakub HR 给出了正确答案。然而，对于我需要 EasyMock 的特殊情况，我可以简单地使用

EasyMock.<List<MyType>>anyObject()

Answer 1

来自 Effective Java:

There are two minor exceptions to the rule that you should not use raw types in new code, both of which stem from the fact that generic type information is erased at runtime. You must use raw types in class literals. The specification does not permit the use of parameterized types (though it does permit array types and primitive types). In other words, List.class, String[].class, and int.class are all legal, but List<String>.class and List<?>.class are not.