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

You don't need to call asScala, it is an implicit conversion:

import scala.collection.JavaConversions._

val javaList = new java.util.LinkedList[String]() // as an example

val scalaList = javaList.iterator.toList

If you really don't have the type parameter of the iterator, just cast it to the correct type:

javaList.iterator.asInstanceOf[java.util.Iterator[String]].toList

EDIT: Some people prefer not to use the implicit conversions in JavaConversions, but use the asScala/asJava decorators in JavaConverters to make the conversions more explicit.