kotlinlang #announcements

does anyone know what would cause a squiggly underline under a pacakage name (folder actually exists) in a class that is specified fully qualified in code? (for example if I were creating a new object from a fully qualified class, e.g.

a.squiggly b.c.d.MyClass()

)

egoscio

09/16/2017, 9:11 AM

I'm getting the following warning, and I've tried googling without any luck:

[kts] kotlin.script.experimental.dependencies.DependenciesResolver.NoDependencies must have a constructor without required parameters

nil2l

09/16/2017, 8:38 PM

Do you mean exception in constructor?

karelpeeters

09/16/2017, 8:50 PM

To actually answer your question, if the parent constructor has at least one parameter you can use this trick to run whatever code you want:

super(param.also { ... })

redrield

09/17/2017, 12:49 AM

What would people recommend for NLP/NLU in Kotlin?

harmony

09/17/2017, 5:46 AM

any eta for kotlin 1.2?

mortda-m

09/17/2017, 3:30 PM

is it possible to write react in kotlin?

fstn

09/17/2017, 5:04 PM

Facebook Chatbot: vertx + java, vertx + scala or vertx + kotlin for a 10 years jee developer?

lovis

09/18/2017, 9:20 AM

the issue here is the name

list

, or

foo

would be equally informative….

neil.armstrong

09/18/2017, 10:55 AM

hmm, I’d probably autogenerate the field name as camelcase then

nil2l

09/18/2017, 12:00 PM

This article is about problems of automatic case mapping. Other is ok.

Let's stop with this nonsense. Either by explicitly declaring the mapping (choice 2) or by avoiding the mapping altogether (choice 3).

mike_shysh

09/18/2017, 12:10 PM

@nil2l he notifies that these annotations just a unnecessary thing, they make a data class less readable. When he need to modify data class, he should manage this field as well

reline

09/18/2017, 1:34 PM

hey guys, I'm having a problem here where when

myString == null

it matches both the

String?

and

Int?

types. Anyone know a way around this?

fun hello(myString: String?, myBoolean: Boolean) {}

fun hello(myInt: Int?, myBoolean: Boolean) {}

fun hi(myString: String?) {
    if (myString == null) {
        hello(myString, false) //overload resolution ambiguity
    } else {
        hello(myString, true) // this works fine
    }
}

anstaendig

09/18/2017, 1:38 PM

Don’t allow myString and myInt to be null in the first place 😎

kevinmost

09/18/2017, 10:00 PM

you're totally allowed to use a lot of the keywords in Kotlin outside of their normal context

orangy

09/18/2017, 10:10 PM

This concept allows evolving language easier in the future. You can’t normally add another hard keyword without breaking some code. E.g. we added

suspend

in 1.1 without breaking any code that might be using

val suspend = shouldSuspendUser()

💯 12

adams2

09/18/2017, 10:52 PM

when would you return null for Unit?

Dalinar

09/19/2017, 6:20 AM

is it ok to override a java default method (in the kotlin subclass interface of the java interface)? I ask because you can't call it via super so was wondering if there could be any unforseen consequences

edwardwongtl

09/19/2017, 8:39 AM

val A.extVal get() = B(this)

Is there a way to not construct

every time I reference to

A.extVal

? Given that I don't own class

Title

edwardwongtl

09/19/2017, 8:39 AM

val A.extVal get() = B(this)

Is there a way to not construct

every time I reference to

A.extVal

? Given that I don't own class

marstran

09/19/2017, 9:06 AM

I tried using the

lazy

-delegate, but it didn't work because you don't have access to

this

inside the function you pass to it. I guess you could create your own version of the

lazy

-delegate that passes the receiver object as a parameter to the function though.

You can read how to do this here: https://kotlinlang.org/docs/reference/delegated-properties.html

This kinda works, but it is not thread safe:

class LazyExtension<V, T>(val f: V.() -> T) {
    data class Result<T>(val t: T)
    var result: Result<T>? = null
    
    operator fun getValue(thisRef: V, property: kotlin.reflect.KProperty<*>): T {
        return result?.t ?: Result(thisRef.f()).also { 
            result = it
        }.t
    }
}

👍 2

edwardwongtl

09/19/2017, 10:22 AM

Wow thanks man, I will take a look

marstran

09/19/2017, 10:49 AM

I fixed it up a little bit so it doesn't use that

tmp

variable.

ilya.gorbunov

09/19/2017, 10:52 AM

@marstran Note that this implementation of LazyExtension would return the same value for any receiver after it had been initialized. Probably not what the author wanted.

marstran

09/19/2017, 10:54 AM

Ah, that's true. You would also need to keep track of all the receivers.

I guess that could turn out to be quite a memory leak if it is used a lot.

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