Is there a Scala equivalent of the Python list unpack (a.k.a. “*”) operator?

Question

In Python, we have the star (or \"*\" or \"unpack\") operator, that allows us to unpack a list for convenient use in passing positional arguments. For example:

Answer 1

You can get some way towards the Python using shapeless,

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scala> import shapeless._
import shapeless._

scala> import Traversables._
import Traversables._

scala> case class ThreeValues(one: String, two: String, three: String)
defined class ThreeValues

scala> val argList = List("one","two","three")
argList: List[String] = List(one, two, three)

scala> argList.toHList[String :: String :: String :: HNil].map(_.tupled).map(ThreeValues.tupled)
res0: Option[ThreeValues] = Some(ThreeValues(one,two,three))

As you can see, a little more ceremony is required in Scala with shapeless. This is because shapeless imposes compile time constraints which are guaranteed to be satisfied at runtime (unlike the python, which will fail at runtime if args is the wrong size or contains elements of the wrong type) ... instead you're forced to specify the type you expect the List to have (in this case exactly three Strings) and be prepared to handle the case where that expectation isn't satisfied (because the result is explicitly an Option of ThreeValues).