Is there a convenient way to initialize a list of strings in Scala?

Question

In Perl I can do

my @l = qw( str1 str2 str3 str4 )

And in Ruby

l = %w{ str1 str2 str3 str4 }

But in Scala it looks like I'm stuck with

val l = List( "str1", "str2", "str3", "str4" )

Do I really need all those "s and ,s?

So how do Perl and Ruby deal with whitespace in str1 for example? — 0__
– 0__, Commented Jan 31, 2013 at 19:19
0__ in perl, "qw" means quote word and splits on white space. For strings with spaces qw is not appropriate, the initialisation would in that case be just like scala — Vorsprung
– Vorsprung, Commented Jan 31, 2013 at 19:24
Yes, what @Vorsprung said. It doesn't cover all cases, but an useful subset. — theglauber
– theglauber, Commented Jan 31, 2013 at 19:48

0__ · Accepted Answer · 2013-01-31 21:31:08Z

16

You could do

implicit class StringList(val sc: StringContext) extends AnyVal {
  def qw(): List[String] = 
    sc.parts.flatMap(_.split(' '))(collection.breakOut)
}

qw"str1 str2 str3"

Or via implicit class:

implicit class StringList(val s: String) extends AnyVal {
  def qw: List[String] = s.split(' ').toList
}

"str1 str2 str3".qw

(Both require Scala 2.10, although the second one can be adapted for Scala 2.9)

edited Jan 31, 2013 at 21:31

answered Jan 31, 2013 at 19:33

0__

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4 Comments

Randall Schulz Over a year ago

Clever use of the new string interpolation factility. It (the first one) is Scala 2.10 only, though.

Daniel C. Sobral Over a year ago

Actually, I'd split the parts, but interpolate args in their original places, without splitting them.

kiritsuku Over a year ago

You must not put arguments to the method. def qw() = ... is enough.

0__ Over a year ago

@sschaef - ah, didn't know. i have edited the answer accordingly

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