java.util.Iterator to Scala list?

Question

I have the following code:

private lazy val keys: List[String] = obj.getKeys().asScala.toList

obj.getKeys returns a java.uti

Answer 1

That would work if obj.getKeys() was a java.util.Iterator. I suppose it is not.

If obj.getKeys() is just java.util.Iterator in raw form, not java.util.Iterator, not even java.util.Iterator, this is something scala tend to dislikes, but anyway, there is no way scala will type your expression as List[String] if it has no guarantee obj.getKeys() contains String.

If you know your iterator is on Strings, but the type does not say so, you may cast :

obj.getKeys().asInstanceOf[java.util.Iterator[String]]

(then go on with .asScala.toList)

Note that, just as in java and because of type erasure, that cast will not be checked (you will get a warning). If you want to check immediately that you have Strings, you may rather do

obj.getKeys().map(_.asInstanceOf[String])

which will check the type of each element while you iterate to build the list