How to create a scala.collection.immutable.Seq from a Java List in Java?

Question

I'm trying to get Akka going in my Java project, and I'm hung up on a small issue with the Seq type(s) from Scala. I'm able to convert my Java List of ActorRef into a scala.collection.Seq, but the Akka API I'm trying to use requires a scala.collection.immutable.Seq. How can I make one?

Code:

static class Router extends UntypedLoadBalancer {
    private final InfiniteIterator<ActorRef> workers;

    public Router(List<ActorRef> workers) {
        Seq workerSeq = asScalaBuffer(workers);

        // how to get from the scala.collection.Seq above to the instance of
        // scala.collection.immutable.Seq required by CyclicIterator below?
        this.workers = new CyclicIterator<ActorRef>();
    }

    public InfiniteIterator<ActorRef> seq() {
        return workers;
    }
}

Landei · Accepted Answer · 2011-07-22 07:17:44Z

48

You can use scala.collection.JavaConversions.asScalaBuffer to convert the Java List to a Scala Buffer, which has a toList method, and a Scala List is a collection.immutable.Seq.

edited Jul 22, 2011 at 7:17

answered Jul 22, 2011 at 6:50

Landei

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2 Comments

spieden Over a year ago

Picked this one as it answered my specific question more closely. The line I ended up with was: new CyclicIterator<ActorRef>((Seq<ActorRef>) asScalaBuffer(workers).toList());

Bryn Over a year ago

as of 2024 the JavaConversion is deprecated, use the scala.jdk.CollectionConverters instead. See the related answer stackoverflow.com/questions/62081253/….

Emil Sit · Accepted Answer · 2011-07-22 02:25:15Z

4

The akka Java documentation for routers as well as the ScalaDoc for CyclicIterator both suggest that the CyclicIterator constructor takes a List.

answered Jul 22, 2011 at 2:25

Emil Sit

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1 Comment

spieden Over a year ago

Thanks for the answer. I guess I was mistaking immutable.Seq for a concrete type!

Dino · Accepted Answer · 2019-11-07 07:51:48Z

3

You can try this:

scala.collection.JavaConverters.asScalaIteratorConverter(list.iterator()).asScala().toSeq();

edited Nov 7, 2019 at 7:51

Dino

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answered Nov 6, 2019 at 20:33

markus

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Comments

Ben McCann · Accepted Answer · 2017-03-12 08:19:22Z

1

You can use:

scala.collection.JavaConverters.collectionAsScalaIterableConverter(workers).asScala().toSeq()

answered Mar 12, 2017 at 8:19

Ben McCann

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Comments

dswiecki · Accepted Answer · 2022-11-05 18:45:32Z

1

Since scala 2.12.0 JavaConversions is marked deprecated

@deprecated("use JavaConverters", since="2.12.0")
object JavaConversions extends WrapAsScala with WrapAsJava

so one should use JavaConverters

scala.collection.JavaConverters.asScalaIteratorConverter(list.iterator()).asScala().toSeq();

or idiomatically

scala.collection.JavaConverters.asScalaIterator(list.iterator()).toSeq();

answered Nov 5, 2022 at 18:45

dswiecki

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