kotlinlang #announcements

I need to interact with a Java-Api, which requires me to extend a class. This class has two different constructors. How can I replicate those two constructors in my class? Kotlin does not allow me to write a secondary constructor that just calls the "secondary" superclass constructor.

nfrankel

05/14/2018, 11:26 AM

you need to chain your own constructors

diesieben07

05/14/2018, 11:27 AM

I can't, I need to call two different super-constructors as they do different things

diesieben07

05/14/2018, 11:29 AM

Found it, can't have a primary constructor in my class.

diesieben07

05/14/2018, 11:29 AM

Then it works.

👍 1

jackjill

05/14/2018, 1:26 PM

🙂

Michael

05/14/2018, 4:22 PM

Did someone say a while a back that it’s “acceptable practice” to just throw java and kotlin into the same

/src/main/java/foo/bar/baz

path?, rather than forking it into

.../java/...

and

.../kotlin/...

👌 1

05/14/2018, 4:23 PM

yes

👍🏼 3

👏 1

rook

05/14/2018, 4:33 PM

I was under the impression that this could impact build times, is that not the case?

fabricio

05/14/2018, 4:36 PM

Is there a List.flatMap with the index ?

karelpeeters

05/14/2018, 4:36 PM

list.withIndex().flatMap { ... }

fabricio

05/14/2018, 4:36 PM

wow.. magic lol

K 5

fabricio

05/14/2018, 4:37 PM

thank you!

tschuchort

05/14/2018, 6:27 PM

Copy code

@Suppress("UNCHECKED_CAST")
fun <X, T : X, S : X> Observable<T>.then(other: Observable<S>): Observable<X>
		= (this as Observable<X>).concatWith(other)

does this look like a terrible idea? It makes the type inference work better

karelpeeters

05/14/2018, 6:30 PM

What observable is this?

tschuchort

05/14/2018, 6:30 PM

what do you mean? any observable

karelpeeters

05/14/2018, 6:30 PM

Observable

is a class, right?

karelpeeters

05/14/2018, 6:30 PM

Where does it come from?

tschuchort

05/14/2018, 6:33 PM

it's from reactivex but doesn't really matter, you can replace it with

List

if you want to. Any monoidal structure will do. The point is that if you take a

List<T>

and a

List<S>

and combine them, the resulting list will have the type of their "closest" common super class

List<X>

karelpeeters

05/14/2018, 6:35 PM

For list that would work because it's covariant.

karelpeeters

05/14/2018, 6:35 PM

But for something like

MutableList

this would be a bad idea.

tschuchort

05/14/2018, 6:36 PM

The same result could be achieved with a function `fun <X> List<X>.combine(other: List<X>)`:

(aList as List<X>).combine(bList)

. But you have to tell the compiler that

aList

is also a

List<X>

manually, which is quite annoying

araqnid

05/14/2018, 10:35 PM

might be easier to make a function that doesn’t depend on Observable/List’s type parameter like that:

fun <X> concat(vararg elements: Observer<X>): Observer<X>

(the Observer class I’m using defines X as having ‘out’ variance) The compiler will choose an X so that all the observers passed as parameters are subtypes

karelpeeters

05/15/2018, 9:55 AM

Where are you querying this?

elect

05/15/2018, 10:03 AM

somewhere later in some init functions

elect

05/15/2018, 10:04 AM

wait

elect

05/15/2018, 10:06 AM

lateinit

here https://github.com/jvm-graphics-labs/Vulkan/blob/master/src/main/kotlin/vulkan/basics/13%20Offscreen%20Rendering.kt#L120 usage here https://github.com/jvm-graphics-labs/Vulkan/blob/master/src/main/kotlin/vulkan/basics/13%20Offscreen%20Rendering.kt#L365

david-wg2

05/15/2018, 11:03 AM

is there a kotlin + bazel hello world example anywhere?