Is it possible to destruct a list into variables e...

# announcements

[supaham]

Is it possible to destruct a list into variables e.g.

val item1, item2, item3 = list

If so, is it possible to make it do a getOrNull?

[adam-mcneilly]

Do you mean, if you try to destructure more items than are in the list, get null?

[supaham]

Well that too. But primarily about item1 = list[0], item2 = list[1], etc.

[supaham]

or more specifically list.getOrNull(0), etc.

[karelpeeters]

val (item1, item2, item3) = list, works up to 5 elements IIRC.

[adam-mcneilly]

hmm well you can destructure the way you have it written, but if you ask for more items than are in the list it won't work.

[supaham]

ahh, alright

[adam-mcneilly]

Also yes, limit of 5 as Karel said

[karelpeeters]

Another idea, create a wrapper object with other componentN functions that return null instead: val (item1, item2, item3) = list.destructOrNull()

[supaham]

whats the method that destructs?

[adam-mcneilly]

Also, if you need more than 5 for your specific use case and really wanna destructure, you can write an extension method for the remaining componentN methods. It becomes a nuisance if you have a lot, but if you wanna go up to 6 you can do this:

operator fun <T> List<T>.component6(): T {
    return this[5]
}

[supaham]

oh, i get it. Thank you!

[adam-mcneilly]

https://kotlinlang.org/docs/reference/multi-declarations.html

[karelpeeters]

destructOrNull would return an instace of something like class ListDestructorWrapper that has those functions defined on it.