Is there a simpler version of zip that doesn t build a seque kotlinlang #getting-started

Channels

100daysofkotlin

100daysofkotlin-2021

algolialibraries

android-architecture

android-databinding

androidgithubprojects

androidx-xprocessing

arrow-contributors

beepiz-libraries

cincinnati-user-group

codemash-precompiler

codingconventions

competitive-programming

competitivecoding

compose-android

compose-desktop

compose-ui-showcase

connect-audit-events

cscenter-course-2016

domain-driven-design

effective-kotlin

effectivekotlin

embedded-kotlin

event21-community-content

fb-internal-demo

fluid-libraries

framework-elide

getting-started

github-workflows-kt

graviton-browser

image-processing

intellij-plugins

intellij-tricks

introduce-yourself

java-to-kotlin-refactoring

jhipster-kotlin

jvm-ir-backend-feedback

k2-early-adopters

kotest-contributors

kotlin-by-example

kotlin-data-storage

kotlin-foundation

kotlin-in-action

kotlin-multiplatform-contest

kotlin-pakistan

kotlinforbeginners

kotlingforbeginners

kotlinprogrammers

kotlintest-devs

kotlinultimatechallenge

kotlinx-datetime

lambdaworld_cadiz

language-evolution

language-proposals

leedskotlinusergroup

library-development

machinelearningbawas

myndocs-oauth2-server

navigation-architecture-component

ohio-kotlin-users

pattern-matching

pocketgitclient

practical-functional-programming

prozis-android-backup

python-contributors

realworldkotlin

refactoring-to-kotlin

refreshversions

russian-kotlinasfirst

skia-wasm-interop-temp

sofl-user-group

spring-security

squarelibraries

stayhungrystayfoolish

strife-discord-lib

swagger-gradle-codegen

swiss-user-group

tricity-kotlin-user-group

Title

m

Mark

05/18/2020, 6:55 AM

Is there a simpler version of zip, that doesn’t build a sequence? Rather, we just do the work in the transform block?

m

marstran

05/18/2020, 6:56 AM

What do you mean? Can you make an example?

m

Mark

05/18/2020, 6:58 AM

seq1.zip(seq2) { (s1, s2) -> doWork(s1, s2) }

So I don’t need the returned sequence

Oh, I still need to iterate over it, of course

k

Kroppeb

05/18/2020, 7:23 AM

I think there is a

zipWith

m

marstran

05/18/2020, 7:32 AM

You could do this:

seq1.zip(seq2).forEach { (s1, s2) ->
  doWork(s1, s2)
}

k

Kroppeb

05/18/2020, 7:36 AM

Oh, I misread the question

m

Mark

05/18/2020, 8:32 AM

@marstran yep, though it’s still creating a new sequence. No big deal though.

m

marstran

05/18/2020, 8:33 AM

Yes, but it's only 1 object creation. It doesn't process anything until

forEach

, like it would if it was a list.

k

Kroppeb

05/18/2020, 8:44 AM

It allocates a bunch of pairs though

m

marstran

05/18/2020, 9:03 AM

@Kroppeb Yeah, true, it does it when consumed. I guess

seq1.zip(seq2) { a, b -> doWork(s1, s2) }.forEach {}

avoids it, but it looks weird 😛

#getting-started Join Slack