如何将 scala.List 转换为 java.util.List? [英] How to convert a scala.List to a java.util.List?

问题描述

如何将Scala的scala.List转换成Java的java.util.List?

解决方案

Scala List 和 Java List 是两种不同的野兽，因为前者是不可变的，后者是可变的.因此，要从一个到另一个，您首先必须将 Scala 列表转换为可变集合.

在 Scala 2.7 上:

import scala.collection.jcl.Conversions.unconvertList导入 scala.collection.jcl.ArrayListunconvertList(new ArrayList ++ List(1,2,3))

从 Scala 2.8 开始:

import scala.collection.JavaConversions._导入 scala.collection.mutable.ListBufferasList(ListBuffer(List(1,2,3):_*))val x: java.util.List[Int] = ListBuffer(List(1,2,3): _*)

但是，如果预期的类型是 Java List，则该示例中的 asList 不是必需的，因为转换是隐式的，如最后一行所示.>

How to convert Scala's scala.List into Java's java.util.List?

解决方案

Scala List and Java List are two different beasts, because the former is immutable and the latter is mutable. So, to get from one to another, you first have to convert the Scala List into a mutable collection.

On Scala 2.7:

import scala.collection.jcl.Conversions.unconvertList
import scala.collection.jcl.ArrayList
unconvertList(new ArrayList ++ List(1,2,3))

From Scala 2.8 onwards:

import scala.collection.JavaConversions._
import scala.collection.mutable.ListBuffer
asList(ListBuffer(List(1,2,3): _*))
val x: java.util.List[Int] = ListBuffer(List(1,2,3): _*)

However, asList in that example is not necessary if the type expected is a Java List, as the conversion is implicit, as demonstrated by the last line.

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