Is it possible to use destructuring in function arguments?

Question

Kotlin supports destructuring declarations:

val (a, b) = Pair(1,2)

This is similar to Python's iterable unpacking:

a, b = (1, 2)

Python also has a splat/spread operator that allows you to perform a similar operation with function arguments:

def f(a, b): pass
pair = (1,2)
f(*pair)

Does kotlin have a similar ability? Obviously, you can unpack the structure manually:

f(pair.component1(), pair.component2())

But that's clunky. Is there a way to do that more elegantly? I don't see anything in the docs on the subject.

user5661440 · Accepted Answer · 2016-03-09 20:01:49Z

19

No, this is possible only for arrays and vararg-functions

val foo = arrayOf(1, 2, 3)
val bar = arrayOf(0, *foo, 4)

answered Mar 9, 2016 at 20:01

user5661440

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1 Comment

AA A Over a year ago

That's too bad. Strange that the spread operator exists in such a limited fashion.

Ingo Kegel · Accepted Answer · 2018-01-14 21:55:25Z

15

Adding to the answer by @Ivan, here are the related issues:

1) spread operator for non-vararg arguments in function calls:

https://youtrack.jetbrains.com/issue/KT-6732

2) destructuring for lambda arguments:

https://youtrack.jetbrains.com/issue/KT-5828

You can vote for them.

Update:

Destructuring for lambda arguments was implemented in Kotlin 1.1.

edited Jan 14, 2018 at 21:55

answered Mar 10, 2016 at 7:30

Ingo Kegel

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Comments

marstran · Accepted Answer · 2016-03-10 11:48:18Z

5

You could define an extension function to spread the arguments of the Pair. Like this:

fun <A, B, R> Pair<A, B>.spread(f: (A, B) -> R) = f(first, second)

fun add(a: Int, b: Int) = a + b

fun main(args: Array<String>) {
    println(Pair(1, 2).spread(::add))
}

This prints 3.

answered Mar 10, 2016 at 11:48

marstran

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1 Comment

nxdrvr Over a year ago

fun <A, B, R> ((A, B) -> R).spread(p:Pair<A, B>) = invoke(p.first, p.second) to pass function which take single pair arg, but actualy two ::myFunc::spread

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