using variable length argument in scala

Question

I know how to define a method with variable length argument:

  case class taxonomy(vocabularies:(String,Set[String])*)

and client code is very clean:

  val terms=taxonomy("topics"->Set("economic","politic")
                   ,"tag"->Set("Libya","evolution")
                   )

but I want to Know how can I use this case class when I have a variable (instead of a Sequence of variable) like this:

val notFormattedTerms = Map("topics"->Set("economic","politic")
       ,"tag"->Set("Libya","evolution"))

Debilski · Accepted Answer · 2011-02-22 15:19:19Z

24

taxonomy(notFormattedTerms.toSeq:_*)

With : _* you virtually transform a sequence argument so that it looks as if a several arguments had been passed to the variable length method. This transformation, however, only works for (ordered?) simple sequence types and, as in this case, not for a Map. Therefore, one will have to use an explicit toSeq before.

edited Feb 22, 2011 at 15:19

answered Feb 22, 2011 at 15:13

Debilski

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2 Comments

Raphael Over a year ago

I have yet to see why these conversions are not done implicitly (where applicable).

Debilski Over a year ago

That would be rather confusing, I suppose.

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