CODEWITHSUNDEEP
javascala
chiappone
chiappone

Reputation: 2908

Whats the equivalent to String... in scala

The title asks the question. Basically I am using a java library that takes a String... as a parameter. How can I call that in my scala code?

I'll add a bit more:

from scala code

def myScalaFunc(params:String*) {
   val myJavaClass = new JavaClass()
   myJavaClass(params)
}

Upvotes: 4

Views: 569

Answers (3)

dhg
dhg

Reputation: 52681

You have to expand the params into a series of arguments, not just a single collection argument. The easiest way to do this is by saying params: _*.

If the Java looks like:

public class VarargTest {
  public void javaFunc(String... args) { 
    // something
  }
}

Then the Scala caller looks like:

def scalaFunc(params: String*) = 
  (new VarargTest).javaFunc(params: _*)

Upvotes: 6

Sean Parsons
Sean Parsons

Reputation: 2832

If the method is defined:

void method(String... param)

Then either call it like this:

method("String 1", "String 2")

Or expand a Seq using this special syntax:

method(Seq("String 1", "String 2"): _*)

Given your example code (I'm assuming my edit is correct):

myJavaClass.myJavaMethod(params: _*)

Upvotes: 5

Óscar López
Óscar López

Reputation: 236024

In Java, this is the syntax for a method that receives a variable number of arguments:

void method(String... param)

In Scala, the equivalent syntax is this:

def method (param:String*)

Upvotes: 1

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