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kotlinlang

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.

<@U0AAQKT9Q> 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