How to create List from Range

Question

I am new to Scala, just started learning, so this is basic beginner question.

I try to implement Sieve of Eratosthenes algorithm. Here is what I got so far:

def sieve_core(cross: Int, lst: Seq[Int]): List[Int] = {
    val crossed = lst.filter(_ % cross != 0)
    crossed match {
            case a :: rest => cross :: sieve_core(a, crossed)
            case _ => cross :: Nil
    }
}

def sieve(max: Int): List[Int] = {
    sieve_core(2, (2 to max))
}

println(sieve(100))

The result is:

List(2)

As far as I understand, case _ => cross :: Nil is matched in first iteration of sieve_core, which means that crossed is not an instance of a List.

I changed lst parameters type to List[Int] and now the code won't compile with an error:

(fragment of Problem3.scala):24: error: type mismatch;
 found   : Range.Inclusive
 required: List[Int]
    sieve_core(2, (2 to max))
                      ^

Apparently Range is not a List.

Question: how can I turn Range into a List? Or is it some bigger problem with my code, I have made some bad assumption somewhere along the way?

Any help appreciated.

Gilles 'SO- stop being evil' · Accepted Answer · 2013-01-30 20:45:45Z

39

There's an apply method on the List companion object which takes a range and returns a List:

scala> List.range(2, 11)
res0: List[Int] = List(2, 3, 4, 5, 6, 7, 8, 9, 10)

There are lots of useful List factory methods in the List collection documentation.

edited Jan 30, 2013 at 20:45

Gilles 'SO- stop being evil'

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answered Sep 11, 2009 at 1:47

DigitalRoss

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5 Comments

Ula Krukar Over a year ago

I checked List class apidocs, haven't checked List object docs. Now I know better. Thanks.

Jonathan Graehl Over a year ago

You know, it's super annoying how they aren't prominently linked to each other, or even combined into a single page :)

Daniel C. Sobral Over a year ago

This is supposed to be deprecated, I'm told. 2 to max toList will work.

DigitalRoss Over a year ago

Thanks, I revised the answer. 2.7.3 docs released in January 2009 gave no hint...

Jonathan Graehl Over a year ago

caution: the very nice option of dropping the "." may still require you to add parens later for methods with no arguments: 2 to 4 toList reverse parses as .toList(reverse), so you need (2 to 4 toList) reverse or (2 to 4).toList reverse or more conventionally, 2.to(4).toList.reverse - but I still drop the ".()" if I can.

Jonathan Graehl · Accepted Answer · 2009-09-11 01:51:10Z

7

To turn any sequence s into a list, use s.toList

I'm sure digitalross' is more efficient in this case, though.

answered Sep 11, 2009 at 1:51

Jonathan Graehl

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Comments

Sal Borrelli · Accepted Answer · 2019-06-17 15:07:41Z

0

(2 to max) is not a scala.collection.immutable.List indeed but a scala.collection.immutable.Range, more precisely an instance of scala.collection.immutable.Range.Inclusive, as mentioned in your error message. Just in passing note that Inclusive and Exclusive are themselves members of Range, with a fairly auto-explanatory meaning.

Luckily the Range class offers the handy method toList, which you can leverage for converting the range into a list and resolve the problem, like in the following code snippet:

scala> val max = 10
max: Int = 10

scala> val r = (2 to max)
r: scala.collection.immutable.Range.Inclusive = Range 2 to 10

scala> val l = r.toList
l: List[Int] = List(2, 3, 4, 5, 6, 7, 8, 9, 10)

answered Jun 17, 2019 at 15:07

Sal Borrelli

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