How to pass Scala array into Scala vararg method?

Question

Consider the code below:

private def test(some:String*){

}

private def call () {
  val some = Array("asd", "zxc")
  test(some)
}

It prints expect String, found Array[String] Why? Are Scala varargs not arrays?

Note

I found several questions on Stack Overflow about Scala varargs, but all of them are about calling Java varargs methods or about converting Scala lists to arrays.

Community · Accepted Answer · 2017-05-23 11:55:10Z

155

Append :_* to the parameter in test like this

test(some:_*)

And it should work as you expect.

If you wonder what that magical :_* does, please refer to this question.

edited May 23, 2017 at 11:55

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answered Jun 26, 2015 at 4:13

Yuhuan Jiang

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Comments

Herrington Darkholme · Accepted Answer · 2015-06-26 04:13:54Z

50

It is simple:

def test(some:String*){}

def call () {
  val some = Array("asd", "zxc")
  test(some: _*)
}

answered Jun 26, 2015 at 4:13

Herrington Darkholme

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Comments

Saurav Sahu · Accepted Answer · 2021-05-23 08:54:46Z

8

Starting Scala 2.13.0, if you use some: _*, you will get this warning:

Passing an explicit array value to a Scala varargs method is deprecated (since 2.13.0) and will result in a defensive copy; Use the more efficient non-copying ArraySeq.unsafeWrapArray or an explicit toIndexedSeq call

As suggested, use scala.collection.immutable.ArraySeq.unsafeWrapArray:

unsafeWrapArray(some):_*

Also, another warning that you should be getting is this one

procedure syntax is deprecated: instead, add : Unit = to explicitly declare test's return type

To fix that, add = just before function's opening bracket:

def test(some:String*) = {

answered May 23, 2021 at 8:54

Saurav Sahu

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