import scala.collection.JavaConverters._ val line: List[String] = null val myTry = Try(line.asJava) val result = myTry match { case Success(_) => "Success" case Failure(_) => "Failure" } println(result) This code snippet prints "Success". If I try to access myTry.get, then it throws a NullPointerException.
From how I understand Try, shouldn't myTry be a Failure?
From how I understand Try, shouldn't myTry be a Failure?
Specifically, asJava on a List creates a wrapper around the original collection in the form of a SeqWrapper. It doesn't iterate the original collection:
case class SeqWrapper[A](underlying: Seq[A]) extends ju.AbstractList[A] with IterableWrapperTrait[A] { def get(i: Int) = underlying(i) } If you use anything else that does iterate the collection or tries to access it, like toSeq, you'll see the failure:
import scala.collection.JavaConverters._ val line: List[String] = null val myTry = Try(line.toSeq) val result = myTry match { case Success(_) => "Success" case Failure(_) => "Failure" } println(result) 
Option[List[String]] instead?Option(line.asJava).isDefined == true. I ended up matching line to case null => null and case _ => line.asJava.Option[T] would be None, not null, and you won't be depending on NullReferenceException, as you wouldn't need a Try[T] at all. You'd only need line.map { x => // stuff }.