Why doesn't Kotlin support unsigned integers?

Question

I came across a situation just recently in which an unsigned integer would have been really useful (e.g. any negative value would not make sense etc.). Surprisingly, I discovered that Kotlin does not support unsigned integers - and there doesn't appear to be anything else out there about why (even though I've looked).

Am I missing something?

https://stackoverflow.com/questions/430346/why-doesnt-java-support-unsigned-ints?rq=1 — nhaarman, May 07 '18 at 19:58
@nhaarman How is a Java answer related to a Kotlin question? — starbeamrainbowlabs, May 07 '18 at 19:59
@starbeamrainbowlabs did you just say what I think you did say? — Vivick, May 07 '18 at 20:25
@Vivick If you're referring to the fact that I don't fully understand how Java and Kotlin are related, then yes. — starbeamrainbowlabs, May 07 '18 at 20:26
Kotlin is transpiled down to Java, which is why there's no "real" support for unsigned integers. You can still hack your way with an `Unsigned` class but you'd encounter several issues. — Vivick, May 07 '18 at 20:29
@Vivick Ah, I see. It all makes sense now! Both why that answer is relevant and why the build system for Kotlin is needlessly complicated...... — starbeamrainbowlabs, May 07 '18 at 20:34
@Vivick, Kotlin is NOT converted to Java source code. It's translated to JVM (Java Virtual Machine) bytecode. Don't disinformate people. — Serid, Jan 31 '19 at 21:45

Willi Mentzel · Accepted Answer · 2021-05-08T11:17:29.253

Unsigned counterparts of Byte, Short, Int and Long do exist in Beta since Kotlin 1.3 and are stable as of Kotlin 1.5:

From the docs:

kotlin.UByte: an unsigned 8-bit integer, ranges from 0 to 255
kotlin.UShort: an unsigned 16-bit integer, ranges from 0 to 65535
kotlin.UInt: an unsigned 32-bit integer, ranges from 0 to 2^32 - 1
kotlin.ULong: an unsigned 64-bit integer, ranges from 0 to 2^64 - 1

Usage

// You can define unsigned types using literal suffixes
val uint = 42u 

// You can convert signed types to unsigned and vice versa via stdlib extensions:
val int = uint.toInt()
val uint = int.toUInt()

score 8 · Answer 2 · answered May 07 '18 at 22:39

8

Why does Kotlin not have native unsigned types

This is because as this question shows, Java does not have built-in unsigned types.

When used on the JVM Kotlin compiles to Java Bytecode, so this limitation on Java also applies to Kotlin.

Workarounds

You can use the utility methods of Integer and Long to operate on values as unsigned link, but this still stores the values as signed internally.

You could also write a utility class that holds a value and acts like an unsigned int type, but this may be slower than using the method above.

answered May 07 '18 at 22:39

jrtapsell

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I don't understand why the JVM dictates the functionality. If you can write the code to imitate an unsigned, then the byte code could do the same. Are you saying the code would be inefficient so the designers just didn't feel it was worth supporting it? It doesn't follow it can't be done. It might follow that it's not worth it. – Mitch Jan 29 '19 at 09:46

score 8 · Answer 3 · answered Sep 06 '18 at 05:55

8

Starting from Kotlin 1.3 unsigned types are available and based on inline classes feature.

See "Unsigned integer types" section of 1.3-M1 release: https://blog.jetbrains.com/kotlin/2018/07/see-whats-coming-in-kotlin-1-3-m1/

answered Sep 06 '18 at 05:55

gildor

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Why doesn't Kotlin support unsigned integers?

3 Answers3

Why does Kotlin not have native unsigned types

Workarounds