Preferred way to create a Scala list

Question

ListBuffer is a mutable list which has constant-time append, and constant-time conversion into a List.

List is immutable and has constant-time prepend and linear-time append.

How you construct your list depends on the algorithm you'll use the list for and the order in which you get the elements to create it.

For instance, if you get the elements in the opposite order of when they are going to be used, then you can just use a List and do prepends. Whether you'll do so with a tail-recursive function, foldLeft, or something else is not really relevant.

If you get the elements in the same order you use them, then a ListBuffer is most likely a preferable choice, if performance is critical.

But, if you are not on a critical path and the input is low enough, you can always reverse the list later, or just foldRight, or reverse the input, which is linear-time.

What you DON'T do is use a List and append to it. This will give you much worse performance than just prepending and reversing at the end.

And for simple cases:

val list = List(1,2,3) 

:)

Uhmm.. these seem too complex to me. May I propose

def listTestD = (0 to 3).toList

or

def listTestE = for (i <- (0 to 3).toList) yield i

You want to focus on immutability in Scala generally by eliminating any vars. Readability is still important for your fellow man so:

Try:

scala> val list = for(i <- 1 to 10) yield i
list: scala.collection.immutable.IndexedSeq[Int] = Vector(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

You probably don't even need to convert to a list in most cases :)

The indexed seq will have everything you need:

That is, you can now work on that IndexedSeq:

scala> list.foldLeft(0)(_+_)
res0: Int = 55