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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?

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    Heh, write a function that swaps two ints :) Yep, switch to C# :-p Commented Apr 27, 2009 at 20:34
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    Mehrdad: Just use Integer rather than int. Commented Apr 27, 2009 at 20:39
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    isn't Integer immutable? Commented Apr 28, 2009 at 0:08
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    @Carlos Heuberger: Yes, but you can get around it with a one-element int[] instead. Commented Apr 28, 2009 at 0:11
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    Wrapping (in Integer or an array) isn't creating pass-by-reference semantics. You're still passing by value; you're just passing a pointer to the wrapper or array. Commented May 22, 2009 at 14:25

6 Answers 6

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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.

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5 Comments

"To prevent confusion, I often mark my method parameters final, which keeps them from being reassigned (accidentally or not) inside the method." I don't believe you, why would you change every single function to make explicit an already existent language feature? Why I ask this is because, one can't really understand Java if it isn't obvious to him that assignment doesn't modify the object.
@Longpoke: I am not the only one who reads my code, so whether or not I understand it isn't really relevant. And I didn't say I changed every single function -- although if you use PMD, which I do, it will warn you every time you don't mark a parameter final.
It's starting to bug me that all variables aren't final by default in Java. @Longpoke--the more restrictive you can be by habit, the less chance of errors. Why would you not want to prevent potential errors whenever possible--even at the "cost" of more explicit code such as creating a new variable instead of re-assigning a parameter? (I personally don't consider more explicit code a cost in most cases, hence the quotes)
@Bill K: Marking parameters as final is okay, if you want to prevent treating the argument as a local variable. But it doesn't make sense to me to mark it final only to "prevent" people from thinking that local argument assignment modifies the outer reference; if they thought that in the first place they probably should go learn Java (or almost any popular OO language?) first. Also, I agree with being restrictive in Java since it sort of goes with the philosophy of static typing, I think I'll mark my parameters final for now on :)
@Longpoke - I think that the real point of confusion is when someone reading the method expects the parameter's value to be the same as when the method call started ... and fails to notice the intervening assignment.
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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.

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2 Comments

The 2 principles of Java variables: 1) variables are passed by value 2) all variables are references (or primitives).
and Java indeed is about the behavior of Objects!
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The references themselves are passed by value.

From Java How to Program, 4th Edition by Deitel & Deitel: (pg. 329)

Unlike other languages, Java does not allow the programmer to choose whether to pass each argument by value or by reference. Primitive data type variables are always passed by value. Objects are not passed to methods; rather, references to objects are passed to methods. The references themselves are passed by value—a copy of a reference is passed to a method. When a method receives a reference to an object, the method can manipulate the object directly.

Used this book when learning Java in college. Brilliant reference.

Here's a good article explaining it. http://www.javaworld.com/javaworld/javaqa/2000-05/03-qa-0526-pass.html

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3 Comments

Excellent description. The most concise and helpful I've read about the subject.
What is the difference between typical passed by reference and reference passed by value? Is there any advantages of doing it or it is just the way Java was design?
If you pass-by-reference in C++, manipulations on the reference actually affect the object that was passed into it. Whereas in Java, even if you assign the reference received as a parameter to the method to a new object, the original object is unchanged because you only received a copy of the reference as a parameter. Take a look at this article (I haven't read the whole thing but the bit about litmus test will interest you): javadude.com/articles/passbyvalue.htm
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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?

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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.

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It's not clear to me what the exact question within the code is. Java is pass-by-value, but arrays are pass-by-reference as they pass no object but only pointers! Arrays consist of pointers, not real objects. This makes them very fast, but also makes them dangerous to handle. To solve this, you need to clone them to get a copy, and even then it will only clone the first dimension of the array.

For more details see my answer here: In Java, what is a shallow copy? (also see my other answers)

By the way, there are some advantages as arrays are only pointers: you can (ab)use them as synchronized objects!

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8 Comments

Java is strictly pass-by-value, even for Arrays! javadude.com/articles/passbyvalue.htm
There are also arrays of non-object types.
@Nico Wawrzyniak Pasting that phrase often enough doesn't make it true.
@Sam Grinrich Correct, but understandig the linked article does.
If you prefer an official source you can read it up in the Java documentation by Oracle, but I find it more difficult to understand compared to the javadude article: docs.oracle.com/javase/specs/jls/se11/html/jls-8.html#jls-8.4.1 and docs.oracle.com/javase/specs/jls/se11/html/…
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