Convert java.util.HashMap to scala.collection.immutable.Map in java

It's entirely possible to use JavaConverters in Java code—there are just a couple of additional hoops to jump through:

import java.util.HashMap;
import scala.Predef;
import scala.Tuple2;
import scala.collection.JavaConverters;
import scala.collection.immutable.Map;

public class ToScalaExample {
  public static <A, B> Map<A, B> toScalaMap(HashMap<A, B> m) {
    return JavaConverters.mapAsScalaMapConverter(m).asScala().toMap(
      Predef.<Tuple2<A, B>>conforms()
    );
  }

  public static HashMap<String, String> example() {
    HashMap<String, String> m = new HashMap<String, String>();
    m.put("a", "A");
    m.put("b", "B");
    m.put("c", "C");
    return m;
  }
}

We can show that this works from the Scala REPL:

scala> val jm: java.util.HashMap[String, String] = ToScalaExample.example
jm: java.util.HashMap[String,String] = {b=B, c=C, a=A}

scala> val sm: Map[String, String] = ToScalaExample.toScalaMap(jm)
sm: Map[String,String] = Map(b -> B, c -> C, a -> A)

But of course you could just as easily call these methods from Java code.

My solution for Java 1.7 and Scala 2.11:

@SuppressWarnings("unchecked")
private static <K, V> scala.collection.immutable.Map<K, V> toScalaImmutableMap(java.util.Map<K, V> javaMap) {
    final java.util.List<scala.Tuple2<K, V>> list = new java.util.ArrayList<>(javaMap.size());
    for (final java.util.Map.Entry<K, V> entry : javaMap.entrySet()) {
        list.add(scala.Tuple2.apply(entry.getKey(), entry.getValue()));
    }
    final scala.collection.Seq<Tuple2<K, V>> seq = scala.collection.JavaConverters.asScalaBufferConverter(list).asScala().toSeq();
    return (scala.collection.immutable.Map<K, V>) scala.collection.immutable.Map$.MODULE$.apply(seq);
}

Since I couldn't make it with Alexey's answer, I'll add that in Scala 2.13 it can be done using:

scala.collection.immutable.Map<Integer, String> scalaMap =
        scala.collection.immutable.Map.from(scala.jdk.CollectionConverters.MapHasAsScala(javaMap).asScala());

This worked for me with java 1.8 and scala 2.12:

public static <K, V> scala.collection.immutable.Map<K, V> toScalaImmutableMap(java.util.Map<K, V> jmap) {
    List<Tuple2<K, V>> tuples = jmap.entrySet()
      .stream()
      .map(e -> Tuple2.apply(e.getKey(), e.getValue()))
      .collect(Collectors.toList());
    
    Seq<Tuple2<K, V>> scalaSeq = JavaConverters.asScalaBuffer(tuples).toSeq();
    
    return (Map<K, V>) Map$.MODULE$.apply(scalaSeq);
}

Since Scala 2.13 you can use CollectionConverters to achieve that.

Having

Map<String, String> javaMap = ...

First we convert it to mutable Scala map:

import scala.jdk.javaapi.CollectionConverters$;

var mutableScalaMap = CollectionConverters$.MODULE$.asScala(javaMap);

And then to immutable one:

var scalaMap = scala.collection.immutable.Map$.MODULE$.from(mutableScalaMap);

Can you provide an additional API call that takes/provides a java.util.Map converted using JavaConverters ?

class Example {
   import scala.collection.JavaConverters._
   def fromMap(m:Map[...]) = ...

   // generics etc. elided
   def fromJava(m:java.util.Map) = {
      fromMap(m.asScala.toMap)
   }
}

You may wish to extract the conversion and provide a decorator (especially as I note you're working to a Scala library). Note dhg's comment re. immutability.

Simply use toMap (Scala 2.12)

import scala.collection.JavaConverters._

// asScala creates mutable Scala Map
// toMap after asScala creates immutable Map

val scalaImmutableMap = javaMap.asScala.toMap