Java Kotlin：相当于 KClass 的 getClass()
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Kotlin: Equivalent of getClass() for KClass
提问by Jire
In Java we can resolve a variable's class through getClass()like something.getClass(). In Kotlin I am aware of something.javaClasswhich is nice but I want to be able to get the KClassin a similar way. I've seen the Something::classsyntax but this is not what I need. I need to get the KClass of a variable. Does such functionality exist?
在 Java 中，我们可以通过getClass()like解析变量的类something.getClass()。在 Kotlin 中，我知道something.javaClass哪个很好，但我希望能够以KClass类似的方式获得。我看过Something::class语法，但这不是我需要的。我需要获取变量的 KClass。是否存在这样的功能？
采纳答案by Alexander Udalov
The easiest way to achieve this since Kotlin 1.1 is the class reference syntax:
自 Kotlin 1.1 以来，实现这一目标的最简单方法是类引用语法：
something::class
If you use Kotlin 1.0, you can convert the obtained Java class to a KClass instance by calling the .kotlinextension property:
如果使用 Kotlin 1.0，则可以通过调用.kotlin扩展属性将获取的 Java 类转换为 KClass 实例：
something.javaClass.kotlin
回答by dherman
EDIT: See comments, below, and answer from Alexander, above. This advice was originally for Kotlin 1.0 and it seems is now obsolete.
编辑：请参阅下面的评论和上面亚历山大的回答。这个建议最初是针对 Kotlin 1.0 的，现在似乎已经过时了。
Since the language doesn't support a direct way to get this yet, consider defining an extension method for now.
由于该语言尚不支持直接获取此方法，因此现在考虑定义扩展方法。
fun<T: Any> T.getClass(): KClass<T> {
return javaClass.kotlin
}
val test = 0
println("Kotlin type: ${test.getClass()}")
Or, if you prefer a property:
或者，如果您更喜欢房产：
val<T: Any> T.kClass: KClass<T>
get() = javaClass.kotlin
val test = 0
println("Kotlin type: ${test.kClass}")
回答by Kashif Anwaar
Here's my solution
这是我的解决方案
val TAG = javaClass.simpleName
With javaClass.simpleName you can obtain your class name. Also the above example is very useful for android developers to declare on top of the class as an instance variable for logging purposes.
使用 javaClass.simpleName 您可以获得您的类名。此外，上面的示例对于 android 开发人员在类的顶部声明为用于日志记录目的的实例变量非常有用。
