How can I get XStream to output Scala lists nicely? Can I write a custom converter?

问题

This code:

println(new XStream.toXML(List(1,2,3)))

produces this XML:

<scala.coloncolon serialization="custom">
  <unserializable-parents/>
  <scala.coloncolon>
    <int>1</int>
    <int>2</int>
    <int>3</int>
    <scala.ListSerializeEnd/>
  </scala.coloncolon>
</scala.coloncolon>

Instead I'd like this:

<list>
  <int>1</int>
  <int>2</int>
  <int>3</int>
</list>

Which would be similar to how the generic java collections get serialized. Whats the best way to do this?

I've got most of the way there by implementing my own converter, but I'm stuck on the unmarshal method, its not clear how to instantiate an empty list...

class ListConverter( _mapper : Mapper )  extends AbstractCollectionConverter(_mapper) {
  /** Helper method to use x.getClass
   * 
   * See: http://scalide.blogspot.com/2009/06/getanyclass-tip.html
   */
  def getAnyClass(x: Any) = x.asInstanceOf[AnyRef].getClass

  def canConvert( clazz: Class[_]) = {       
    classOf[::[_]] == clazz
  }

  def marshal( value: Any, writer: HierarchicalStreamWriter, context: MarshallingContext) = {
    val list = value.asInstanceOf[List[_]]
    for ( item <- list ) {      
      writeItem(item, context, writer)
    }
  }

  def unmarshal( reader: HierarchicalStreamReader, context: UnmarshallingContext ) = {
    println(context.getRequiredType())
    var list : List[_] = createCollection(context.getRequiredType()).asInstanceOf[List[_]]
    while (reader.hasMoreChildren()) {
      reader.moveDown();
      val item = readItem(reader, context, list);
      list = item :: list
      reader.moveUp();
    }
    list
  }
}

object ListConverter {
  def configureXStream( stream: XStream ) = {
    stream.alias("list", classOf[::[_]])
    stream.registerConverter( new ListConverter(stream.getMapper) )        
  }
}

回答1:

Not seconds after posting the question, the answer came to me, here is a working implementation of unmarshal:

  def unmarshal( reader: HierarchicalStreamReader, context: UnmarshallingContext ) = {
    var list : List[_] = Nil 
    while (reader.hasMoreChildren()) {
      reader.moveDown();
      val item = readItem(reader, context, list);
      list = list ::: List(item) // be sure to build the list in the same order
      reader.moveUp();
    }
    list
  }

回答2:

There is only one instance of an empty list, which is the object Nil.

来源：https://stackoverflow.com/questions/1753511/how-can-i-get-xstream-to-output-scala-lists-nicely-can-i-write-a-custom-convert

标签

generics

list

scala

xstream