Convert from scala.collection.Seq to java.util.List in Java code

Question

I'm calling a Scala method, from Java. And I need to make the conversion from Seq to List.

I can't modified the signature of the Scala method, so I can't used the asJavaCollection method from scala.collection.JavaConversions._

Any ideas of how can I achieve this?

Using Scala 2.9.3

Chris Martin · Accepted Answer · 2019-04-20T22:07:41.893

48

You're on the right track using JavaConversions, but the method you need for this particular conversion is seqAsJavaList:

java.util.List<String> convert(scala.collection.Seq<String> seq) {
    return scala.collection.JavaConversions.seqAsJavaList(seq);
}

Update: JavaConversions is deprecated, but the same function can be found in JavaConverters.

java.util.List<String> convert(scala.collection.Seq<String> seq) {
    return scala.collection.JavaConverters.seqAsJavaList(seq);
}

edited Apr 20 '19 at 22:07

answered Jul 19 '13 at 04:25

Chris Martin

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I'm not able to **import scala.collection.JavaConversions;** in eclipse. its showing Error here – Ashish Ratan Mar 20 '14 at 05:43
Not sure if this is new or not but I had to do `seqAsJavaList(seq).asJava` to get the actual `List` – grinch Mar 04 '15 at 19:41
Just added an anonymous edit to this point to use the new JavaConverters class (since 2.8.1). Hopefully will be approved. – Decoded Apr 20 '19 at 19:36
@Decoded: Thanks for the improvement :) I didn't accept the edit as-is because I wanted the answer to still make sense in the context of the question, but I added an amendment that uses `JavaConverters`. – Chris Martin Apr 20 '19 at 22:09
Thanks! Love to be on the cutting edge :D – Decoded Apr 21 '19 at 03:25

score 8 · Answer 2 · edited Sep 18 '14 at 17:47

8

Since Scala 2.9, you shouldn't use implicits from JavaConversions since they are deprecated and will soon be removed. Instead, to convert Seq into java List use convert package like this (although it doesn't look very nice):

import scala.collection.convert.WrapAsJava$;

public class Test {
    java.util.List<String> convert(scala.collection.Seq<String> seq) {
        return WrapAsJava$.MODULE$.seqAsJavaList(seq);
    }
}

edited Sep 18 '14 at 17:47

Mark

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answered Jul 19 '13 at 07:06

4lex1v

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It might be a good idea for scala to provide a more java-friendly set of frontend methods for these kinds of conversions. – Rüdiger Klaehn Jul 19 '13 at 08:12

score 8 · Answer 3 · answered Jul 18 '17 at 20:36

Since 2.12 this is the recommended way:

public static <T> java.util.List<T> convert(scala.collection.Seq<T> seq) {
    return scala.collection.JavaConverters.seqAsJavaList(seq);
}

All other methods a @deprecated("use JavaConverters or consider ToJavaImplicits", since="2.12.0")

rasen58 · Answer 4 · 2017-08-11T00:07:54.983

0

(In case you want to do this conversion in Scala code)

You can use JavaConverters to make this really easy.

import collection.JavaConverters._
val s: Seq[String] = ...
val list: java.util.List<String> = s.asJava

edited Aug 11 '17 at 00:07

answered Aug 01 '17 at 02:07

rasen58

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The question asks for the conversion in Java code, not Scala code. – Tom Wang Aug 10 '17 at 23:22
Ah my mistake, I knew I found it strange that no one had suggested this solution yet - makes sense now. – rasen58 Aug 11 '17 at 00:07

Convert from scala.collection.Seq to java.util.List in Java code

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