Converting mutable collection to immutable

Question

I'm looking for a best way of converting a collection.mutable.Seq[T] to collection.immutable.Seq[T].

Answer 1

As specified:

def convert[T](sq: collection.mutable.Seq[T]): collection.immutable.Seq[T] =    collection.immutable.Seq[T](sq:_*) 

Addition

The native methods are a little tricky to use. They are already defined on scala.collection.Seq and you’ll have to take a close look whether they return a collection.immutable or a collection.mutable. For example .toSeq returns a collection.Seq which makes no guarantees about mutability. .toIndexedSeq however, returns a collection.immutable.IndexedSeq so it seems to be fine to use. I’m not sure though, if this is really the intended behaviour as there is also a collection.mutable.IndexedSeq.

The safest approach would be to convert it manually to the intended collection as shown above. When using a native conversion, I think it is best practice to add a type annotation including (mutable/immutable) to ensure the correct collection is returned.

Answer 2

If you want to convert ListBuffer into a List, use .toList. I mention this because that particular conversion is performed in constant time. Note, though, that any further use of the ListBuffer will result in its contents being copied first.

Otherwise, you can do collection.immutable.Seq(xs: _*), assuming xs is mutable, as you are unlikely to get better performance any other way.