Java is NEVER pass-by-reference, right?...right? [duplicate]

Question

Possible Duplicate:
Is Java “pass-by-reference”?

I found an unusual Java method today:

private void addShortenedName(ArrayList<String> voiceSetList, String vsName)
{
     if (null == vsName)
       vsName = "";
     else
       vsName = vsName.trim();
     String shortenedVoiceSetName = vsName.substring(0, Math.min(8, vsName.length()));
     //SCR10638 - Prevent export of empty rows.
     if (shortenedVoiceSetName.length() > 0)
     {
       if (!voiceSetList.contains("#" + shortenedVoiceSetName))
         voiceSetList.add("#" + shortenedVoiceSetName);
     }
}

According to everything I've read about Java's behavior for passing variables, complex objects or not, this code should do exactly nothing. So um...am I missing something here? Is there some subtlety that was lost on me, or does this code belong on thedailywtf?

Answer 1

As Rytmis said, Java passes references by value. What this means is that you can legitimately call mutating methods on the parameters of a method, but you cannot reassign them and expect the value to propagate.

Example:

private void goodChangeDog(Dog dog) {
    dog.setColor(Color.BLACK); // works as expected!
}
private void badChangeDog(Dog dog) {
    dog = new StBernard(); // compiles, but has no effect outside the method
}

Edit: What this means in this case is that although voiceSetList might change as a result of this method (it could have a new element added to it), the changes to vsName will not be visible outside of the method. To prevent confusion, I often mark my method parameters final, which keeps them from being reassigned (accidentally or not) inside the method. This would keep the second example from compiling at all.

Answer 2

Java passes references by value, so you get a copy of the reference, but the referenced object is the same. Hence this method does modify the input list.