Java - How to convert Scala Stream to Java Stream?

Question

I have a Java application that receives a Scala Stream.

I need to convert this to a Java Stream.

How can I do this in Java?

It's not clear what you're asking. Which exactly type you have on Scala side, and which type you need on Java side? — Haspemulator
– Haspemulator, Commented Nov 25, 2015 at 16:22
Scala side is creating a Scala Stream, and on the Java side I need to convert this to a Java input/output stream, — jon lee
– jon lee, Commented Nov 25, 2015 at 18:00

Xavier Guihot · Accepted Answer · 2022-12-22 07:37:28Z

6

Starting Scala 2.13, the standard library includes scala.jdk.javaapi.StreamConverters which provides Java to Scala implicit stream conversions:

import scala.jdk.javaapi.StreamConverters;

// val scalaStream = Stream(1, 2, 3)
StreamConverters.asJavaSeqStream(scalaStream);
// java.util.stream.Stream[Int] = java.util.stream.ReferencePipeline$Head@3cccf515

Note that Streams are renamed LazyList in Scala 2.13, in which case:

// val scalaLazyList = LazyList(1, 2, 3)
StreamConverters.asJavaSeqStream(scalaLazyList);
// java.util.stream.Stream[Int] = java.util.stream.ReferencePipeline$Head@4997c13

edited Dec 22, 2022 at 7:37

answered Mar 30, 2019 at 15:09

Xavier Guihot

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Comments

lyjackal · Accepted Answer · 2015-11-26 03:38:33Z

This is a bit of a long way around, but you can convert the scala stream to an iterable, to a java iterable, then construct the java stream from the java iterable:

scala>
import java.util.stream.StreamSupport
import scala.collection.JavaConverters._

def stream2javaStream[T](scalaStream: scala.Stream[T]): java.util.stream.Stream[T] = {
    StreamSupport.stream(scalaStream.toIterable.asJava.spliterator(), false);
}


stream2javaStream((1 to 100).toStream)

res0: java.util.stream.Stream[Int] = java.util.stream.ReferencePipeline$Head@2489e84a

While roundabout, this does not "realize" the stream, maintaining its efficiency.

scala>
stream2javaStream((1 to 100).toStream.map{i => println(i); i})
1
res1: java.util.stream.Stream[Int] = java.util.stream.ReferencePipeline$Head@9b21bd3

only prints 1 for the head of the stream

robbo · Accepted Answer · 2018-10-30 22:51:35Z

If you don't want to process the stream in parallel you can simply subclass AbstractSpliterator:

import java.util.function.Consumer
import java.util.stream.StreamSupport
import java.util.{Spliterator, Spliterators}

def streamToJava[T](s: Stream[T]): java.util.stream.Stream[T] = {
    var ss: Stream[T] = s
    val splitr = new Spliterators.AbstractSpliterator[T](Long.MaxValue, Spliterator.IMMUTABLE) {
      override def tryAdvance(action: Consumer[_ >: T]): Boolean = ss match {
        case Stream.Empty => false
        case h #:: tl => {ss = tl; action.accept(h); true}
      }
    }
    StreamSupport.stream(splitr, false)
  }

The other answer (I'm too new to comment) is incorrect as it calls size which materialises the entire stream.

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