Hey I m wondering if there is an operator on `Flow` that ach kotlinlang #coroutines

that achieves the following: If the Flow has not produced at least one value in a given timeframe (x miliseconds), produce a default value followed by any values the Flow might produce afterwards. In contrast to a solution with

debounce

however, if the Flow has produced a value before the timeframe, the default value will just be ignored and the value will be emitted immediatly.

Something like a

delayedDefault

😉

I built something myself but I'm not sure if this is the best solution 😁

fun <T> Flow<T>.delayedDefault(
    defaultAfter: Long,
    defaultValue: () -> T
): Flow<T> =
    channelFlow {
        val job = launch {
            delay(defaultAfter)
            offer(defaultValue())
        }

        collect {
            job.cancel()
            offer(it)
        }
    }

elizarov

01/20/2020, 9:45 AM

Looks good to me. However, it would be safer to use

send(defaultValue())

just in case somebody does

delayedDefault(…).buffer(0)

👍🏼 1

svenjacobs

01/20/2020, 9:45 AM

Okay, thanks for the feedback from the expert 👍🏼 😉

elizarov

01/20/2020, 9:47 AM

You should also add one more

job.cancel()

after collect, so that it does not wait in the corner case of empty flow (

flowOf().delayedDefault(…)

)

svenjacobs

01/20/2020, 9:47 AM

Oh, right!

elizarov

01/20/2020, 9:47 AM

Unless you want delay+default in this case, too

svenjacobs

01/20/2020, 9:48 AM

Should I also use

send()

inside

collect

elizarov

01/20/2020, 9:48 AM

Yes, that, too

svenjacobs

01/20/2020, 9:49 AM

Cancelling an already cancelled

Job

will do nothing, right? No exception will be thrown?

elizarov

01/20/2020, 9:50 AM

Yes. It is perfectly fine and it is quite fast, too.

👍🏼 3

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