Can t find a more elegant way to do this ``` fun foo vararg kotlinlang #announcements

Can't find a more elegant way to do this. ``` fun ...

hastebrot

05/23/2016, 4:44 PM

Can't find a more elegant way to do this.

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fun foo(vararg bar: String) = bar.joinToString(" ")

foo(*listOf("inky", "blinky").toTypedArray())

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