Scala convert List[Int] to a java.util.List[java.lang.Integer]

Apparently you need both conversions. However, you can group them in a single implicit conversion:

implicit def toIntegerList( lst: List[Int] ) =
  seqAsJavaList( lst.map( i => i:java.lang.Integer ) )

Example:

scala> def sizeOf( lst: java.util.List[java.lang.Integer] ) = lst.size

scala> sizeOf( List(1,2,3) )
res5: Int = 3

Because the underlying representation of Int is Integer you can cast directly to java.util.List[java.lang.Integer]. It will save you an O(n) operation and some implicit stuff.

import collection.JavaConversions._

class A {
  def l() = asList(List(1,2)).asInstanceOf[java.util.List[java.lang.Integer]]
}

Then you can use from Java like this:

A a = new A();
java.util.List<Integer> l = a.l();

Note that on 2.9.0 ,I get a deprecation warning on asList (use seqAsJavaList instead)

Did you try:

val javalist = collection.JavaConversions.asJavaList (y)

I'm not sure, whether you need a conversion Int=>Integer or Int=>int here. Can you try it out?

Update: The times, they are a changing. Today you'll get a deprecated warning for that code. Use instead:

import scala.collection.JavaConverters._
val y = List (1)
> y: List[Int] = List(1)

val javalist = (y).asJava
> javalist: java.util.List[Int] = [1]