Java is NEVER pass-by-reference, right?...right???

Question

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

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.

Answer 2

Well, it can manipulate the ArrayList - which is an object... if you are passing an object reference around (even passed by value), changes to that object will be reflected to the caller. Is that the question?

Answer 3

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 4

I think you are confused because vsName is modified. But in this context, it is just a local variable, at the exact same level as shortenedVoiceSetName.

Answer 5

See my article, "Java is Pass-by-Value Dammit!"

http://javadude.com/articles/passbyvalue.htm