Scala: Is there a difference between method parameter String* and Array[String]

Question

Is there a difference between a method parameter String* and Array[String]?

Console:

scala> def main(args: Array[String]): Unit = {}
main: (args: Array[String])Unit

scala> def main(args: String*): Unit = {}
main: (args: String*)Unit

Code 1:

object Example {
  def main(args: Array[String]): Unit = {
    println("Hello")
  }
}

>> Hello

Code 2:

object Example {
  def main(args: String*): Unit = {
    println("Hello")
  }
}

>> Exception in thread "main" java.lang.NoSuchMethodException: Example.main([Ljava.lang.String;)
    at java.lang.Class.getMethod(Class.java:1786)
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:126)

Possible duplicate of How to pass scala Array into scala vararg method? — prayagupadhyay
– prayagupadhyay, Commented Aug 30, 2017 at 23:40

puhlen · Accepted Answer · 2017-08-30 23:15:59Z

7

Yes,

String* is a varargs, it takes in any number of Strings, which will be passed to the method as a Seq[String].

Array[String] take in a single Array of Strings.

If you have a sequence of String that you want to pass as a String*, you cannot do so directly, but you can "splat" out the sequence to pass it in using the :_* type ascription.

def varArg(input: String*){}

val strings = Seq("hello", "world")
varArg(strings:_*)

answered Aug 30, 2017 at 23:15

puhlen

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4 Comments

rapt Over a year ago

So is String* a type or just how varargs is expressed in the console? How can I see in the console that it's passed to the method as Seq[String]? P.S. This old answer says it is passed as Array, did it change at some point to Seq? stackoverflow.com/a/1438796

puhlen Over a year ago

@rapt I think it is a type but I'm not sure. It is only usable as a method parameter either way. The more upvoted and detailed answer form that question also mentions that it is a Seq. You can also find out for yourslef in the repel by defining a varargs method and seeing what you get. def test(args: String*) = args. We see that it returns a Seq[String], the actual implementation depends though, when we pass in arguments the concrete implementation is WrappedArray but if we pass no arguments it is Nil (the empty List aka. List())

som-snytt Over a year ago

I downvoted the linked answer for that error; Java uses arrays, Scala uses Seq. The repeated param has a type that is a Seq, scala-lang.org/files/archive/spec/2.12/…. The vararg itself is a type internal to the compiler, but we don't normally see that.

som-snytt Over a year ago

It would be cool if you could get the compiler to make a bridge to a varargs main, I don't know why they don't do that. Worth adding that this is what App is for.

som-snytt · Accepted Answer · 2017-08-31 05:25:19Z

2

I went and asked on the forum.

The annotation you want is annotation.varargs.

I had some inkling that they had that glue, but I never had a reason to use it.

Personally, I'd rather it were automatic for a main method, b/c obvious.

$ scala
Welcome to Scala 2.12.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_131).
Type in expressions for evaluation. Or try :help.

scala> object Main { @annotation.varargs def main(args: String*) =
     | println(args.size)
     | }
defined object Main

scala> Main.main(Array("hello","world"): _*)
2

OK, cool.

answered Aug 31, 2017 at 5:25

som-snytt

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1 Comment

som-snytt Over a year ago

Probably my last input on SO.

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