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?
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())
@Suppress("PLATFORM_CLASS_MAPPED_TO_KOTLIN")
Pattern#split is an optional way, I'll fix my answer. thanks for your feedback, sir. How about it now?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.
