Kotlin: Equivalent of getClass() for KClass

Question

In Java we can resolve a variable's class through getClass() like something.getClass(). In Kotlin I am aware of something.javaClass which is nice but I want to be able to get the KClass in a similar way. I've seen the Something::class syntax but this is not what I need. I need to get the KClass of a variable. Does such functionality exist?

Alexander Udalov · Accepted Answer · 2017-06-19T12:48:36.373

229

The easiest way to achieve this since Kotlin 1.1 is the class reference syntax:

something::class

If you use Kotlin 1.0, you can convert the obtained Java class to a KClass instance by calling the .kotlin extension property:

something.javaClass.kotlin

edited Jun 19 '17 at 12:48

answered Sep 18 '15 at 15:39

Alexander Udalov

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Any place I can follow the implementation of something::class? – Nishad Feb 11 '16 at 18:21
1

@Nishad The easiest place would be to look at the generated bytecode (e.g. with `javap`) – Alexander Udalov Feb 12 '16 at 08:36
2

You can also use `KClass::class` to get the current class object. – Saminda Peramuna May 16 '18 at 06:10
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Something to note is that `something::class` returns `KClass` while `something.javaClass.kotlin` returns `KClass` – KylePlusPlus Jul 18 '18 at 15:58
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@SamindaPeramuna This is just wrong, KClass::class returns the class object kotlin.reflect.KClass not the current class object! To get it you have to use this::class. – Markus Kreusch Jan 07 '20 at 09:54
Just use ```something::class.simplename``` – Raj Narayanan Feb 03 '20 at 15:20

dherman · Answer 2 · 2018-07-10T22:35:14.457

14

EDIT: See comments, below, and answer from Alexander, above. This advice was originally for Kotlin 1.0 and it seems is now obsolete.

Since the language doesn't support a direct way to get this yet, consider defining an extension method for now.

fun<T: Any> T.getClass(): KClass<T> {
    return javaClass.kotlin
}

val test = 0
println("Kotlin type: ${test.getClass()}")

Or, if you prefer a property:

val<T: Any> T.kClass: KClass<T>
    get() = javaClass.kotlin

val test = 0
println("Kotlin type: ${test.kClass}")

edited Jul 10 '18 at 22:35

answered Apr 13 '16 at 05:14

dherman

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You can skip the `java` bit and go straight Kotlin: `fun T.getClass(): KClass = this::class` – TWiStErRob May 14 '18 at 19:00
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This answer should be updated to reflect the changes in 1.1. Now it is possible to get the class using `someClass::class` – Max Jul 10 '18 at 03:41

score 12 · Answer 3 · answered Apr 22 '20 at 15:14

12

Here's my solution

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.

answered Apr 22 '20 at 15:14

Kashif Anwaar

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score 3 · Answer 4 · answered Feb 07 '23 at 12:14

3

I think this is the better solution. Found here https://www.techiedelight.com/determine-class-name-in-kotlin/

fun main() {

val s = "Kotlin"

println(s::class)                           // class kotlin.String
println(s::class.qualifiedName)             // kotlin.String
println(s::class.simpleName)                // String

}

answered Feb 07 '23 at 12:14

user2164852

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As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Feb 09 '23 at 05:18

score 2 · Answer 5 · answered Jun 15 '20 at 11:27

Here are different Implementations to get class names. You can utilize it as per your requirements.

import kotlin.reflect.KClass

val <T : Any > T.kClassName: KClass<out T>
get() {
    return javaClass.kotlin
}

Here we can get the class name in kotlin

val <T : Any > T.classNameKotlin: String?
get() {
    return javaClass.kotlin.simpleName
}

Here we can get the class name in kotlin

val <T : Any > T.classNameJava: String
get() {
    return javaClass.simpleName
}

Here are the outputs to the following operations.

fun main(){

val userAge = 0

println(userAge.kClassName) 
Output: class java.lang.Integer (Kotlin reflection is not available)

println(userAge.classNameKotlin)
Output: Int

println(userAge.classNameJava)
Output: Integer
}

score 2 · Answer 6 · answered Aug 25 '21 at 11:55

Since Kotlin 1.5.21 (org.jetbrains.kotlin:kotlin-stdlib:1.5.21)

val TAG = javaClass.simpleName will work no more!

Error "Not enough information to infer type variable T"

/**
 * Returns the runtime Java class of this object.
 */
public inline val <T : Any> T.javaClass: Class<T>
    @Suppress("UsePropertyAccessSyntax")
    get() = (this as java.lang.Object).getClass() as Class<T>

@Deprecated("Use 'java' property to get Java class corresponding to this Kotlin class or cast this instance to Any if you really want to get the runtime Java class of this implementation of KClass.", ReplaceWith("(this as Any).javaClass"), level = DeprecationLevel.ERROR)
public inline val <T : Any> KClass<T>.javaClass: Class<KClass<T>>
    @JvmName("getRuntimeClassOfKClassInstance")
    @Suppress("UsePropertyAccessSyntax")
    get() = (this as java.lang.Object).getClass() as Class<KClass<T>>

You can try now YourClassName::class.java.simpleName

score 1 · Answer 7 · edited Feb 25 '21 at 20:28

1

In Kotlin:

val className = serviceClass.javaClass.name

edited Feb 25 '21 at 20:28

β.εηοιτ.βε

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answered Feb 25 '21 at 17:46

Diseño Web Cantabria

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Kotlin: Equivalent of getClass() for KClass

7 Answers7