CODEWITHSUNDEEP
kotlin
Renato
Renato

Reputation: 13690

How can I use a Java String method (split) directly from Kotlin source?

Doing some profiling, it seems my bottleneck is the Kotlin kotlin.CharSequence.split() extension function.

My code just does something like this:

val delimiter = '\t'
val line = "example\tline"
val parts = line.split(delimiter)

As you might notice, parts is a List<String>.

I wanted to benchmark using Java's split directly, which returns a String[] and might be more efficient.

How can I call Java's String::split(String) directly from Kotlin source?

Upvotes: 3

Views: 1924

Answers (2)

holi-java
holi-java

Reputation: 30686

You can cast a kotlin.String to a java.lang.String then use the java.lang.String#split since kotlin.String will be mapped to java.lang.String , but you'll get a warnings. for example:

//                               v--- PLATFORM_CLASS_MAPPED_TO_KOTLIN warnings
val parts: Array<String> = (line as java.lang.String).split("\t")

You also can use the java.util.regex.Pattern#split instead, as @Renato metioned it will slower than java.lang.String#split in some situations. for example:

val parts: Array<String> = Pattern.compile("\t").split(line, 0)

But be careful, kotlin.String#split's behavior is different with java.lang.String#split , for example:

val line: String = "example\tline\t"

//   v--- ["example", "line"]
val parts1: Array<String> = Pattern.compile("\t").split(line, 0)

//   v--- ["example", "line", ""]
val parts2 = line.split("\t".toRegex())

Upvotes: 5

s1m0nw1
s1m0nw1

Reputation: 81929

You can do this:

(line as java.lang.String).split(delimiter)

But it's not recommended to not use the kotlin.String as the compiler might tell you.

Upvotes: 2

Related Questions