如何创建已知类型的类文字:Class<List<String>> [英] How to create a class literal of a known type: Class&lt;List&lt;String&gt;&gt;

问题描述

采取以下措施:

public Class<List<String>> getObjectType() {
    // what can I return here?
}

我可以从这个方法返回什么样的类文字表达式来满足泛型并编译?List.class 不会编译，List.class 也不会编译.

What class literal expression can I return from this method which will satisfy the generics and compile? List.class won't compile, and neither will List.<String>class.

如果您想知道为什么"，我正在编写 Spring 的 FactoryBean<List<String>> 的实现，这需要我实现 Class>getObjectType().然而，这不是一个 Spring 问题.

If you're wondering "why", I'm writing an implementation of Spring's FactoryBean<List<String>>, which requires me to implement Class<List<String>> getObjectType(). However, this is not a Spring question.

我悲痛的呼声已经被 SpringSource 的权力听到了，所以 Spring 3.0.1 将把 getObjectType() 的返回类型更改为Class，巧妙地避免了这个问题.

edit: My plaintive cries have been heard by the powers that be at SpringSource, and so Spring 3.0.1 will have the return type of getObjectType() changed to Class<?>, which neatly avoids the problem.

推荐答案

你总是可以投射到你需要的，像这样

You can always cast to what you need, like this

return (Class<List<String>>) new ArrayList<String>().getClass();

或

return (Class<List<String>>) Collections.<String>emptyList().getClass();

但我认为这不是您想要的.好吧，它可以工作，但有警告，但它并不完全漂亮".

But I assume that's not what you are after. Well it works, with a warning, but it isn't exactly "beautiful".

我刚找到这个

为什么通配符参数化类型没有类字面量?

Why is there no class literal for wildcard parameterized types?

因为通配符参数化类型没有精确的运行时类型表示.

Because a wildcard parameterized type has no exact runtime type representation.

所以投射可能是唯一的出路.

So casting might be the only way to go.

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