How to convert a Java Stream to a Scala Stream?

Question

As a part of an effort of converting Java code to Scala code, I need to convert the Java stream Files.walk(Paths.get(ROOT)) to Scala. I can't find a solution by googling. asScala won't do it. Any hints?

Here is the related code:

import static org.springframework.hateoas.mvc.ControllerLinkBuilder.linkTo;
import static org.springframework.hateoas.mvc.ControllerLinkBuilder.methodOn;

import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.stream.Collectors;

// ...snip...

Files.walk(Paths.get(ROOT))
     .filter(path -> !path.equals(Paths.get(ROOT)))
     .map(path -> Paths.get(ROOT).relativize(path))
     .map(path -> linkTo(methodOn(FileUploadController.class).getFile(path.toString())).withRel(path.toString()))
     .collect(Collectors.toList()))

The Files.walk(Paths.get(ROOT)) return type is Stream<Path> in Java.

Answer 1

There is a slightly nicer way without needing the compat layer or experimental 2.11 features mentioned here by @marcospereira

Basically just use an iterator:

import java.nio.file.{Files, Paths}
import scala.collection.JavaConverters._

Files.list(Paths.get(".")).iterator().asScala

Answer 2

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

import scala.jdk.StreamConverters._

val javaStream = Files.walk(Paths.get("."))
// javaStream: java.util.stream.Stream[java.nio.file.Path] = java.util.stream.ReferencePipeline$3@51b1d486
javaStream.toScala(LazyList)
// scala.collection.immutable.LazyList[java.nio.file.Path] = LazyList(?)
javaStream.toScala(Iterator)
// Iterator[java.nio.file.Path] = <iterator>

Note the usage of LazyList (as opposed to Stream) as Streams are deprecated in Scala 2.13. LazyList is the supported replacement type.