[英]Return a Class instance with its generic type
Here's a simple example that demonstrates a type-erasure-related issue I am running into. 这是一个简单的示例,演示了我遇到的与类型擦除相关的问题。 I have a class like this:
我有一个这样的课:
public abstract class AbstractHandler<T> {
...
public abstract Class<T> handledType();
}
Then I have this implementation: 然后我有这个实现:
public class ConcreteHandler extends AbstractHandler<Map<String, List<Thing>>> {
@Override
public Class<Map<String, List<Thing>>> handledType() {
//what do I return here?!
}
}
I can't return Map<String, List<Thing>>.class
, since that's not even valid syntactically. 我不能返回
Map<String, List<Thing>>.class
,因为它甚至不能在语法上有效。 I tried making the generic type-parameter in the subtype to be HashMap<String, List<Thing>>
and then returning new HashMap<String, List<Thing>>().getClass()
, but that doesn't work because the return type of Class<T>#getClass()
is Class<? extends T>
我尝试将子类型中的泛型类型参数设置为
HashMap<String, List<Thing>>
然后返回new HashMap<String, List<Thing>>().getClass()
,但这不起作用,因为返回类型Class<T>#getClass()
是Class<? extends T>
Class<? extends T>
. Class<? extends T>
。 I looked at TypeToken
from Guava, and the getRawType
method seemed promising, but it returns Class<? super T>
我从Guava看了
TypeToken
, getRawType
方法看起来很有希望,但它返回Class<? super T>
Class<? super T>
. Class<? super T>
。
I have a workaround for the time being that looks like this: 我现在有一个解决方法,看起来像这样:
public class ThingListMap {
private Map<String, List<Thing>> thingListMap;
...
}
and I just use ThingListMap
as the generic type-parameter. 我只是使用
ThingListMap
作为泛型类型参数。
Another possible workaround is to perform a forced cast: 另一种可能的解决方法是执行强制转换:
public Class<Map<String, List<Thing>>> handledType() {
return (Class<Map<String, List<Thing>>>) new HashMap<String, List<Thing>>().getClass();
}
Is there a more-elegant way to do this? 有更优雅的方式吗?
EDIT: In response to one of the answers, I cannot change the signature of the handledType
method since I do not own or control its source. 编辑:响应其中一个答案,我无法更改
handledType
方法的签名,因为我不拥有或控制其来源。
For some reason, Java doesn't allow you to cast Map.class
directly to Class<Map<String, List<Thing>>>
. 出于某种原因,Java不允许您将
Map.class
直接转换为Class<Map<String, List<Thing>>>
。 It's an unchecked cast anyway. 无论如何,这是一个未经检查的演员。
But it's legal to cast it twice, first to Class<?>
, then to Class<Map<String, List<Thing>>>
. 但是将它强制转换两次是合法的,首先是
Class<?>
,然后是Class<Map<String, List<Thing>>>
。
return (Class<Map<String, List<Thing>>>) (Class<?>) Map.class;
Being an unchecked cast, you may want to add @SuppressWarnings("unchecked")
. 作为未经检查的演员,您可能想要添加
@SuppressWarnings("unchecked")
。
Guava's approach to this is to use TypeToken
s. Guava的方法是使用
TypeToken
。 Your class would become 你的课将成为
public abstract class AbstractHandler<T> {
public TypeToken<T> handledType();
}
public class ConcreteHandler extends AbstractHandler<Map<String, List<Thing>>> {
@Override
public TypeToken<Map<String, List<Thing>>> handledType() {
return new TypeToken<Map<String, List<Thing>>>() {};
}
}
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