将Scala Seq转换为Java列表

Question

I am working in Java. I am calling a scala function which returns a Seq[String]. To convert it to java's List, I tried using scala.collection.JavaConverters's asJava. But this does not seem to work.

Answers on similar questions have suggested either using JavaConversions or WrapAsJava, both of which are deprecated.

Similar Question - Converting Scala seq<string> to Java List<string>

//someScalaFunc returns a Seq[String]
List<String> listA = someScalaFunc(); 

Answer 1

由于JavaConversions 已被弃用 ，你可以使用JavaConverters此：

List<String> listA = scala.collection.JavaConverters.seqAsJavaList(someScalaFunc())

Answer 2

您可以通过导入以下代码来使用Java到Scala隐式转换器：

import scala.collection.JavaConversions._

Answer 3

import scala.collection.JavaConverters._

val scalaList = List("A", "B", "c")

scalaList.asJava