Kotlin语言在运行时获取类 [英] Kotlin language get class at runtime

问题描述

假设我们有以下内容:

val person = "Bill"

有人可以解释一下两者之间的区别吗?

Could someone explain the difference between these two:

val kClass1 = person.javaClass.kotlin    

与

val kClass2 = person::class

什么时候我应该打一个而不是另一个?

When I should call the one instead of the other?

任何源代码示例都将不胜感激.

Any source code example would be appreciated.

推荐答案

有两种方法可以实现相同的目的，即获取对象的Kotlin类，是因为在Kotlin 1.1之前，::class文字不支持其左侧的表达式.因此，如果您使用的是Kotlin 1.0，则唯一的选择是.javaClass.kotlin，否则您都可以使用它们.这就是"Kotlin in Action"使用.javaClass.kotlin语法的原因:它是在Kotlin 1.1发行版之前编写的.

The main reason there are two ways to achieve the same thing, namely get the Kotlin class of an object, is because prior to Kotlin 1.1, the ::class literal did not support the expression on its left-hand side. So if you're using Kotlin 1.0, your only option is .javaClass.kotlin, otherwise you're fine with either of them. This is the reason "Kotlin in Action" uses the .javaClass.kotlin syntax: it was written before the Kotlin 1.1 release.

这些表达式的类型也略有不同.例如，在以下代码中

There's also a minor difference in the types of these expressions. For example, in the following code

interface T

fun f1(x: T) = x::class
fun f2(x: T) = x.javaClass.kotlin

f1的类型为KClass<out T>，但是f2的类型为KClass<T>.这实际上是对javaClass声明的疏忽:在这种情况下，KClass<out T>更正确，因为x的类不一定是T，但也可以是T的子类.

f1's type is KClass<out T>, but f2's type is KClass<T>. This is actually an oversight in the javaClass declaration: KClass<out T> is more correct in this case, because x's class is not necessarily T, but can also be a subclass of T.

否则，这两个表达式(x.javaClass.kotlin和x::class)在产生的字节码和运行时性能方面是完全等效的.我更喜欢x::class，因为它更短并且读起来更好.

Otherwise these two expressions (x.javaClass.kotlin and x::class) are completely equivalent in terms of produced bytecode and runtime performance. I prefer x::class because it's shorter and reads better.

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