Another question: this code snipped:

fun doStuff(input:String) {
   val split = input.split("foo", 0)
}

will not compile, but this one:

fun doStuff(input:String) {
   val split = java.lang.String(input).split("foo", 0)
}

will be OK. Is it supposed behaviour (in other words, is it a bug or is it a feature)?

thats expected behaviour, because

kotlin.String.split

is not the same as

java.lang.String.split

, they have different parameters

You can use

val split = input.split("foo", limit = 0)

. Also mind that: 1)

limit = 0

is the default value for

limit

, so there's no need to specify it explicitly. 2)

split("foo")

treats

foo

delimiter as a literal rather than a regular expression.

@ilya.gorbunov Yes, I saw the Kotlin version of

split

, and I found that behaviour of

limit

is different with Java. Java doc says that

If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.

, however Kotlin implementation keeps trailing empty strings