Java is Pass by Value and Not Pass by Reference

Filed Under: Interview Questions

One of the biggest confusion in Java programming language is whether java is Pass by Value or Pass by Reference. I ask this question a lot in interviews and still see the interviewee confused with it. So I thought to write a post about it to clear all the confusion around it.

First of all, we should understand what is meant by pass by value or pass by reference.

  • Pass by Value: The method parameter values are copied to another variable and then the copied object is passed, that’s why it’s called pass by value.
  • Pass by Reference: An alias or reference to the actual parameter is passed to the method, that’s why it’s called pass by reference.

Update: I am getting a lot of comments and it seems that still there is a lot of confusion, so I have made a video tutorial to explain this in detail.

Java is always Pass by Value and not pass by reference, we can prove it with a simple example.

Let’s say we have a class Balloon like below.

package com.journaldev.test;

public class Balloon {

	private String color;

	public Balloon(){}
	public Balloon(String c){
	public String getColor() {
		return color;

	public void setColor(String color) {
		this.color = color;

And we have a simple program with a generic method to swap two objects, the class looks like below.

package com.journaldev.test;

public class Test {

	public static void main(String[] args) {

		Balloon red = new Balloon("Red"); //memory reference 50
		Balloon blue = new Balloon("Blue"); //memory reference 100
		swap(red, blue);
		System.out.println("red color="+red.getColor());
		System.out.println("blue color="+blue.getColor());
		System.out.println("blue color="+blue.getColor());

	private static void foo(Balloon balloon) { //baloon=100
		balloon.setColor("Red"); //baloon=100
		balloon = new Balloon("Green"); //baloon=200
		balloon.setColor("Blue"); //baloon = 200

	//Generic swap method
	public static void swap(Object o1, Object o2){
		Object temp = o1;

When we execute above program, we get following output.

red color=Red
blue color=Blue
blue color=Red

If you look at the first two lines of the output, it’s clear that swap method didn’t worked. This is because Java is pass by value, this swap() method test can be used with any programming language to check whether it’s pass by value or pass by reference.

Let’s analyze the program execution step by step.

Balloon red = new Balloon("Red");
Balloon blue = new Balloon("Blue");

When we use new operator to create an instance of a class, the instance is created and the variable contains the reference location of the memory where the object is saved. For our example, let’s assume that “red” is pointing to 50 and “blue” is pointing to 100 and these are the memory location of both Balloon objects.

Now when we are calling swap() method, two new variables o1 and o2 are created pointing to 50 and 100 respectively.

So below code snippet explains what happened in the swap() method execution.

public static void swap(Object o1, Object o2){ //o1=50, o2=100
	Object temp = o1; //temp=50, o1=50, o2=100
	o1=o2; //temp=50, o1=100, o2=100
	o2=temp; //temp=50, o1=100, o2=50
} //method terminated

Notice that we are changing values of o1 and o2 but they are copies of “red” and “blue” reference locations, so actually there is no change in the values of “red” and “blue” and hence the output.

If you have understood this far, you can easily understand the cause of confusion. Since the variables are just the reference to the objects, we get confused that we are passing the reference so java is passed by reference. However, we are passing a copy of the reference and hence it’s pass by value. I hope it clears all the doubts now.

Now let’s analyze foo() method execution.

private static void foo(Balloon balloon) { //baloon=100
	balloon.setColor("Red"); //baloon=100
	balloon = new Balloon("Green"); //baloon=200
	balloon.setColor("Blue"); //baloon = 200

The first line is the important one, when we call a method the method is called on the Object at the reference location. At this point, balloon is pointing to 100 and hence it’s color is changed to Red.

In the next line, ballon reference is changed to 200 and any further methods executed are happening on the object at memory location 200 and not having any effect on the object at memory location 100. This explains the third line of our program output printing blue color=Red.

I hope the above explanation clears all the doubts, just remember that variables are references or pointers and its copies are passed to the methods, so java is always pass-by-value. It would be more clear when you will learn about Heap and Stack memory and where different objects and references are stored, for a detailed explanation with reference to a program, read Java Heap vs Stack.


  1. ccdan says:

    “Java is always Pass by Value and not pass by reference, we can prove it with a simple example” – totally WRONG!

    If java is “always Pass by Value” then when you pass an object to a method, you should have an ENTIRE COPY OF THAT OBJECT – as it happens in C++, and the method should operate solely on that COPY, not on the original!

    Objects in java are ALWAYS PASSED BY REFERENCE(which technique always uses a copy of the address where the original variable points to – people are confounding this with pass-by-value)

    the “swap” example is totally irrelevant as it doesn’t deal with “pass by reference” in any way, but it relies on the ability of a variable to point to another variable name, which is not permitted in java

  2. Zhe says:

    i think this is inaccurate statement, the ‘pass by value’ in java not always fall into your definition, for example, array, when you pass the array into a function it the copy of the reference to the underlying array that is passed into the function. so if you change the array the change is going to be reflected outside of the function call as well.

  3. Zeynel says:

    I think you are wrong,

    Method parameters stores reference addresses. How can you call it pass by value while it is storing references.

  4. bb says:

    public static void swap(Object o1, Object o2){
    Object temp = o1;

    o1 and o2 are local variables inside of the swap method’s scope. You just swapped the value of two local variables, why would it cause effect on the objects itself? You didn’t do anything with the 2 object.

    1. Tom Goldman says:

      this is the best and easiest explanation of the problem

  5. B Senthil Kumaran says:

    You have clearly given the addresses copied in the function parameters too which is same as the called time address value. Hence, its passed by reference only.

    If it is passed by value, the value alone will be copied and the address will not be the same. Hence, the foo(Balloon balloon) function first line will not work if it copied by value alone.

    Since, it’s pass by reference, the modification happened on the location from ‘Blue’ to ‘Red’ reflected in the main() function where it is called.

    Balloon red = new Balloon(“Red”); //memory reference 50
    Balloon blue = new Balloon(“Blue”); //memory reference 100

    System.out.println(“blue color=”+blue.getColor());

    private static void foo(Balloon balloon) { //baloon=100
    balloon.setColor(“Red”); //baloon=100
    balloon = new Balloon(“Green”); //baloon=200
    balloon.setColor(“Blue”); //baloon = 200

    The problem here is, since the Java is not clear in call by reference (not exactly the pointer, the real method of handling the objects through addresses like in C & CPP) and pass by value, the new address assigned is not getting reflect in the called time variable’s / object’s address.

    Where it failed to copy/update the new address assigned in the called function to the original address. Somehow it seems like bug but, its the design actually and if you need to get the new addressed value, you have to get it on return value and assign in the calling area.

  6. kamlesh parmar says:

    thanks for sharing, explanation is very clear.

    1. priya says:

      Its so clear. Thanks

  7. Senecaek2 says:

    if you have an array it is Pass by Reference … “Java is always Pass by Value” … seems to be wrong

    1. Fandazky says:

      Can you give explanation or example for ” array it is Pass by Reference ” ?
      because i just try with array and i think array also pass by value

  8. Ronak Vora says:

    This is awesome blog and i have increased my knowledge regarding java from this, thank you very much for writing this. Keep writing.

  9. levan says:

    this article causes more confusion than explain anything clearly at all..
    :))) you say objects are copied before passed down? thats wrong… object are not, but their references are copied, objects stays the same
    that means, one can not dereference variable inside method because its a copy, BUT, and thats a big BUT that same copy points to same object as original reference so if you modified object it will be modified everywhere its used.

    1. Suhas says:

      Hi Levan,

      Watch this video by Microsoft Learning-
      Am sure it will clarify all your doubts.

      1. architect says:

        Suhas, great tip. Excellent and concise video really explains the issue. This post, however, no matter how well intended, is not correctly worded, thus very confusing.

  10. someone says:

    I think the confusion is about relativity.

    If you look at the object, then we can say it is passed by reference.
    If you look at the variable, then we can say it is passed by value (whatever it is, being an int, float or a reference towards an object).

    Method arguments are local variables anyways, of course they are passed by value.

    It is important to define the “what” that is passed to a method first. Seems evident to someone who got the subtility, not to target audience.

    It is not about Java vs C/C++, it is the same passing behaviour. The only difference is that in C/C++, you can have object variables and not only references towards objects, meaning that you pass an object by value. But a pointer or reference stays a variable which is passed by value, even if you use a pointer to a pointer to … to an object.

    1. theRiley says:

      Here is the way I understand the concept. For primitives, Java is strictly pass-by-value. For objects, Java is pass-reference-by-value. Since the reference is strong, operations on members of objects passed succeed. Since the reference is a copy, or value, operations on the references themselves are simply variables on a method call stack which do not effect the references in the calling process.

  11. Jeff says:

    If you say Java is pass by value, you must base it against a lanuage like C++ where the definition is widely accepted, nuanced, and understood.
    When you create your balloons in your sample Java code, you are:
    – creating a pointer on the stack, which points to an object on the heap.
    – creating a new frame upon calling the swap method, which copies the object arguments (pointers) into it’s own stack space. note now our heap address is copied inside both the main stack frame and the swap stack frame.
    – reassign the swap stack frame’s local variables and their contents

    You can do the same thing in C++:
    – create a pointer on the stack, pointing to memory on the heap:
    Balloon * red = new Balloon(“red”);
    Balloon * blue = new Balloon(“blue”);
    swap(red, blue);
    – creating a new stack in the swap method, which copies the object arguments (pointers) into it’s own stack space
    swap(Balloon * r, Ballon * b) {
    Balloon * temp = r;
    r = b;
    b = temp;
    – in c++, you can print out specific memory addresses. If you did so in the main method, you’d find that red and blue would be low heap addresses, 0x0235897 for example. If you dereference, &red and &blue, you’d find they’re high stack addresses, 0x7ffff235 for example.
    In the swap method, r and b are copy constructored so while r and b do refer to the same low heap addresses as red and blue, &r and &b are different addresses in the stack as they’re not in the same stack as the original, nor are they the same pointer (like in pass by reference).

    So really, if you’re using pass-by-value, pass-by-reference, pass-by-pointer terminology the way C++ does, java is pass by pointer for objects, and pass by value for primitives.

  12. Paweł says:

    I understand how Java is passing the parameters but I don’t agree that it is passing by value.
    Look at your definition:
    “Pass by Reference: An alias or reference to the actual parameter is passed to the method, that’s why it’s called pass by reference.”
    That is exactly what Java does. It is coping a reference to object and placing a copy on a stack memory with other parameters. There is no deep copy of the object itself (which we usually name as Pass by value).
    In C++ you can also pass by pointer. The pointer is copied to the stack as any other parameters of the method. The fact that we copy value of the pointer does not mean that we pass parameter by value.

    1. Jeff says:

      exactly. java does for objects exactly what C/C++ does when you pass-by-pointer.
      the point of confusion lies in that you’re passing the pointer to the object, but passing the value of the pointer. the two really mean the same thing. but it’s contrary to convention to say that you’re passing by value here when the value is a memory address – that’s just playing with words and making up your own definitions that, while don’t contradict standard english definitions, certainly goes against convention in our techne.

  13. Jonathan says:

    Pass by reference is actually possible in Java though.

    If for example, you have a method that takes an interface like this.

    public static void passByRefExample(List foo)

    And then you invoke it using an object with allocated heap space. Like so.

    public static void main(String[] args)
    List foo = new ArrayList();
    passByRefExample(foo); -> System.out.println(bar));


    1. Jonathan says:

      Those are \ lists btw, my carats got eaten by the formatting bunny.

      1. Jonathan says:

        String lists even

    2. Jonathan says:

      Another example would be if we did something along these lines.

      public interface BalloonInterface {
      public String getColor();
      public void setColor(String color);

      public class Balloon implements BalloonInterface{
      private String color;

      public String getColor(){
      return color;

      public void setColor(String color){
      this.color = color;

      //So now what can we do? Well, let’s make a method to test this out.

      public static void swapColor(BalloonInterface balloon){

      //then we can invoke it in main like this.
      public static void main(String[] args)
      BalloonInterface balloon = new Balloon();


  14. Dim Gab says:

    Try this modified swap function:
    public static void swap(Balloon balloon, Balloon balloon2){
    Balloon temp = balloon;

    for more confusion 🙂

    1. Amine says:

      your swap function is not actually swapping the objects, but just calling a method in the referenced object.
      There is a difference ….

  15. Vivek says:

    I think it is perfect explanation..My concept is cleared….!!!
    It is just that a new reference copied reference variable is passed in the arguments..

  16. baozi says:

    thanks !!! that settled my doubts

  17. Electrical Engineering Student says:

    Thank you for this explanation and the video. I am an electrical engineering student, but take computer science classes as part of my curriculum. My professor taught us Java was pass-by-value, and our textbook said the same but I found those explanations a little confusing because the net effect of some methods really seem to be pass-by-reference based on my understanding of it. Your foo method is a good example of this.

    However, after reading your article, my class notes, and the relative sections in my book many times, I am finally starting to understand. Also, your video was exceptionally helpful by going over stack and heap memory. When it clicked for me is when I realized the value stored in the reference variable (which has its own memory address) is the memory address of the object. Therefore, the value passed to a method is a copy of the object’s memory address. Though a method can mutate the object at the received memory address, it cannot mutate the value of the address stored in the original reference variable, and so the language is pass-by-value.

    Thanks again!

  18. Honza Zidek says:

    It is a good interview queston if you are talking with a senior programmer who knows multiple languages and understands the Pascal / C terminology.
    On other side, if you talk with a pure Java programmer, it is rather an evil catch, because Java (misuses) the term “reference” for something which in Pascal / C corresponds with (but is not equivalent to) “pointer”.

    The correct answer is e.g. here and the exact wording should be like this

    > Java is always pass-by-value. When we pass the value of an object, we are passing the reference to it.

  19. patient-zero says:

    The reference of the object is passed by value

  20. Java Developer says:

    I totally disagree with this article.
    Java is pass by reference that’s why we are unable to swap two objects.
    Reference is a pointer when you assign a new object to it then the reference starts pointing to the new object. and you can’t change the original object.
    In the foo function, color is changed to red and in the main function, color red is printed. Because the object was passed by reference.
    You can never explain how this can happen in call by value.

    1. ComputerScientist says:

      A reference is not a pointer. A pointer is a type. A reference is alias to an l-value which cannot be null or changed.

  21. Akash says:

    Great !!

    red color in o1=Blue
    blue color in o2=Red
    red color=Red
    blue color=Blue
    blue color=Red

    This output would have cleared the confusion more effectively.

  22. Nikola says:

    This article is good if you have some previous understanding of Java passing. Basically, you proved nothing here. Why? Well, you said that Java is passed by value, not by ref. But in the case of swap(), you are passing object reference. And the reason why swapping is not successful is because you did all the swap thing inside of the method which basically didn’t have any effect on base objects. You were just switching memory location inside a method by adding it to a temp object variable. Real switch/update of the object you did within the foo method by updating ballon from blue to red.

    Please, don’t confuse us 🙂

    Primitive types are passed by value but objects are pass by reference.

  23. Stephane says:

    This article achieved an almost impossible feast. Turning muddy water into mud. I’m kidding. 🙂

    You assume making a copy of the reference and passing it to the method is a pass by value, because a copy is passed. That is not what I and many understand. By doing so you confuse me and all these many people.

    But I want to thank you anyway !

    First because you incited me to write this answer:

    And second because I like your articles very much.

  24. Wagner says:

    I think a C programmer will easily understand this.
    If we want to swap pointers, we need to pass the pointer of pointers. *p
    If we want to swap values, we need to pass the pointer of values. *val

  25. Rajat Mehla says:

    If Java is pass by value and not reference, how come Collection.sort(ArrayListName) sorts the collection object, when we are not re-assigning the value to original object? We are just calling sort method with reference of collection.

    1. Shervin says:

      We are not re-assigning it, but check the sort() method in Collection. It is re-assigning.

  26. Curious George says:

    Sorry, minute 6:40, not 7:45.

    1. Pankaj says:

      If you think in terms of String Pool, then yes I should just point it to Red object. However, I wanted to keep it simple and not mix the string pool concept with pass by value concept. It could have been any other type of custom object too.

      1. Curious George says:

        I understand. Thank you!

  27. Curious George says:


    thanks for your work. It’s very good.

    Around minute 7:50, though, could you have just erased blue’s arrow and redrawn it pointing to red, or did you intentionally change the box from blue to red?

  28. alireza alallah says:

    I think it’s wrong , just swap reference address object

  29. Rob says:

    I think a lot of the confusion stems from the verbiage. Expressions like, “pass by value” and “pass by reference” lead people to think we are, “passing the value” and “passing the reference”, respectively. It’s clearer to simply say, “we are not passing the value or the reference, we are passing a NEW reference to the same value.”

    1. Curious George says:

      Excellent summary. Thank you. Adds to the video explanation, which I found very good.

  30. Mayank Kumar says:

    Can you explain the array behaviour here :

    public class PassByValueTest {

    public static void main(String[] args) {
    PassByValueTest p = new PassByValueTest();

    private void start() {
    long[] a1 = { 3, 4, 5};
    long[] a2 = fix(a1);
    System.out.println(a1[0]+” “+a1[1]+” “+a1[2]+” “);
    System.out.println(a2[0]+” “+a2[1]+” “+a2[2]);
    boolean b1 = false;
    boolean b2 = fix(b1);
    System.out.println(b1+” “+b2);
    int a = 5;
    int b = fix(a);
    System.out.println(a+” “+b);

    private int fix(int c) {
    return c;

    private boolean fix(boolean b1) {
    b1 = true;
    return b1;

    private long[] fix(long[] a3) {
    a3[1] = 7;
    return a3;

    output :
    3 7 5
    3 7 5
    false true
    5 7
    according to pass by value concept first line should be 3 4 5 kindly explain this behaviour.

    1. Anonymous says:

      In Java, all primitives like int, char, etc are similar to C/C++, but all non-primitives (or objects of any class) are always references. So it gets tricky when we pass object references to methods. Java creates a copy of references and pass it to method, but they still point to same memory reference. Mean if set some other object to reference passed inside method, the object from calling method as well its reference will remain unaffected.

  31. Noe says:

    Java is pass by value ok but careful : if you pass an Object as parameter you are passing the reference to it. So whatever you do with your parameter inside the method will actually modify the Object passed. Of course the reference will not be changed.

    1. mp77 says:

      You’re still passing a value (copy) of the reference, NOT the reference itself. You can assign new object to that copied reference and outter reference will still point to the old object!

  32. robert says:

    Hi Pankaj,
    Thank you for your article and video. The video helped. I was still a bit confused though and tried your swap code and added my own swap2 method to call the set method explicitly on the objects o1 and o2. The effect of this does in fact change the values in the main method. So although you say it is “pass by value” what we are seeing is a change in value taking place on the called objects because we still have a reference to the objects passed to the method.

    public class PassbyValueExample {

    public static void main(String args []){

    Balloon red = new Balloon(“red”);
    Balloon blue = new Balloon(“blue”);

    swap(red, blue);

    System.out.println(“ex1. red colour = ” + red.getColour());
    System.out.println(“ex1. blue colour = ” + blue.getColour());

    swap2(red, blue);

    System.out.println(“ex2. red colour = ” + red.getColour());
    System.out.println(“ex2. blue colour = ” + blue.getColour());


    private static void swap(Object o1, Object o2){
    Object temp = o1;
    o1 = o2;
    o2 = temp;

    private static void swap2(Object o1, Object o2){
    Object temp = o1;
    o1 = o2;
    o2 = temp;



    1. Satish says:

      package headfirst.designpatterns.adapter.ducks;

      public class PassbyValueExample {

      public static void main(String args[]) {

      Balloon red = new Balloon(“red”);
      Balloon blue = new Balloon(“blue”);

      swap(red, blue);

      System.out.println(“ex1. red colour = ” + red.getColour());
      System.out.println(“ex1. blue colour = ” + blue.getColour());

      swap2(red, blue);

      System.out.println(“ex2. red colour = ” + red.getColour());
      System.out.println(“ex2. blue colour = ” + blue.getColour());


      private static void swap(Object o1, Object o2) {
      Object temp = o1;
      o1 = o2;
      o2 = temp;

      private static void swap2(Object o1, Object o2) {
      // Object temp = o1;
      // o1 = o2;
      // o2 = temp;

      // o1 and o2 are copies of red and blue objects and these copies sent as values
      // and hence Pass by value
      // So in caller method swap2(red, blue), red and blue are references to actual
      // objects created in heap and in called method(current method) o1 and o2 are
      // copies of the red and blue which point to same object in heap
      // As o1 and o2 are also references for actual objects you can modify object
      // like below
      ((Balloon) o1).setColour(“blue”);
      ((Balloon) o2).setColour(“red”);

      // After below 2 statements o1 and o2 pointing to new objects in heap memory.

      o1 = new Balloon(“red”);
      o2 = new Balloon(“blue”);

      System.out.println(“ex2. red colour = ” + ((Balloon) o1).getColour());
      System.out.println(“ex2. blue colour = ” + ((Balloon) o2).getColour());

      * The above change only effect in current method as o1 and o2 scope is only in
      * swap2 method and after swap2 method finished jvm only know red and blue
      * references which still point to old objects If Java is Pass by reference then
      * after swap2 call o1 and o2 color should be red and blue.


  33. Mubashar says:

    The reason people are confused is because in this blog and many other definition of pass-by-value and pass-by-reference has been changed. probably because its coming from a who never considered C++ probably mother of all modern languages while having this discussion.

    Actually your foo example is exactly tells us Java is pass-by-reference since as per C++ perspective. if swap method would have worked it would mean java is pass-by-reference-of-reference. In c++ this is referred as double pointer (**). Now let me write definitions.

    pass-by-value: If you pass in an object and callee changes any value it would not be available because when caller called the method application cloned a new object in the stack memory and pass in the calee.

    pass-by-reference: If you pass in an object and callee changes any value it would be available because when caller called the method application provided the value of reference of the same object so callee can change anything but one exception it can’t call on “new” on the same reference since the reference passed by value.

    To overcome this issue C++ provides double pointer (**) operator in this case you pass the reference of a reference to an object. in such case you can modify it as well as you can use new operator all together.

    In java there is no direct way available but workarounds are to use the wrapper classes.

    1. 7thArcanist says:

      I’m sorry but you got it wrong. I recommend watching this video ( to understand the situation.

      Also, this thread on StackOverflow may be useful:

  34. lala says:

    If somebody wrote something like this in a job interview, I’d politely tell her/him the interview is finished.

    //Generic swap method
    public static void swap(Object o1, Object o2){
    Object temp = o1;

    1. Thiru says:

      why dude? whats wrong? How long you have been coding in Java?

      1. Alex says:

        Because it doesn’t achieve a swap.

  35. Parth says:

    Thanks for the video tutorial. It really cleared the concept. Thanks a lot!!

  36. ramason says:

    I’m Java Developer in Korea 🙂
    Can I upload the post to my blog translating this?
    The another code ‘ll be added of the post!
    If you’ll allow, please answer me.
    Thank you.

    1. Pankaj says:

      Rather than commenting here, it would be good if you contact me directly. Usually we don’t allow republishing of our articles elsewhere.

  37. Progstacks says:

    I would say it is both pass-by-reference and pass-by-value and somewhat confusing. I did a litter experiment and I posted in my blog.

  38. karatedog says:

    You all fighting terminology here. Look at Assembly, where everything is a value, period. Any other fancy name (like “reference” or “pointer”) is the way you use this value.
    Semantically, the sentence: “A copy of a reference is value” is false, If you changed “reference” to “dog”, and “value” to “cat” it is probably easier to see.
    A copy of a pointer is, well… could be a pointer. Or my mother’s birthday, if it happens to be exactly 1941.
    So Java is pass by value, because someone long ago stated, that a copy of a reference is a value. That’s all.
    Anyway, check Stackoverflow for Is Java “pass-by-reference” or “pass-by-value”?, the question is there since 2012.

  39. Lokesh says:

    Hello Pankaj,

    If you think the program is passing balloon objects as pass by value. Consider this program, it is swapping the values now.

    package com.baloon;

    public class Test {

    public static void main(String[] args) {

    Balloon red = new Balloon(“Red”); //memory reference 50
    Balloon blue = new Balloon(“Blue”); //memory reference 100

    swap(red, blue);
    System.out.println(“red color=”+red.getColor());
    System.out.println(“blue color=”+blue.getColor());


    //Generic swap method
    public static void swap(Balloon o1, Balloon o2){
    /* Object temp = o1;
    o2=temp; */


    Output :

    red color=Blue
    blue color=Red

  40. David says:

    Do you really ask this question in job interviews? These are the type of questions that make common interview practices as useful as a coin flip.

    Java passes variables by value (where object non primitive variables are actually references to an object). If someone can demonstrate they know how this works, rather than trying to answer a question that is known to be interpreted different ways. Have you ever known someone where answering this was the proving factor on their success? I guarantee there are people that have read that java passes object references by value, but would not be able to tell you the output of your tests.

    Please stop asking questions like this. Everyone, please stop using interview questions like this. Asking “is java pass by value or pass by reference” when you know people are going to get confused looks on their faces only serves the purpose of feeding ones ego an sense of supiority. Instead try seeing if the candidate can actually write correct functioning code, that they can grok what code examples may do. More importantly, put an effort in the interview to determine if the candidate is bright, driven and a strong personality fit for your team. This will improve your rate of hiring strong team members. Itwill reduce the number of false negatives which come from candidates incorrectly answering pointed trivia questions that help the interviewer feel good about themselves. It will also reduce the number of missed job opportunities for candidates who were asked poor questions that prove nothing about potential for success.

    The interview is. It about you (the I terviewer) it is about the candidate. Don’t leave an interview feeling smart because you were able to trick a candidate. Instead, leave the interview feeling good because you helped the candidate really show what the have to offer.

    Sorry to rant – I get frustrated with people perpetuating poor interview practices for the ‘80s. I know there are other good articles here that help provide deeper understanding of java. However, this article just feels disingenuous and a little smug to me.

    1. Pankaj says:

      Hi David,

      If you have faced or taken a lot of interviews, you must know that it’s not only about giving correct answers. Nobody expects to get 100% correct answers and any good interviewer don’t judge if an answer is wrong. Some questions are confusing and they are still asked to see how interviewee responds to it, how he reacts when follow-up or counter questions are asked.

      Technical knowledge can get you only half way, confidence, way of talking, problem solving approach, soft skills play a vital role in getting a job.

      The article is written to help you understand the topic clearly, there is nothing “disingenuous and a little smug” part in it.

  41. Himanshu says:

    Appreciate your vedio

  42. William L. DeRieux IV says:

    I believe that you don’t fully understand pass by value and pass by reference.

    primitive parameters are passed by value.
    reference parameters are passed by reference.

    pass by value means a copy of the primitive is sent to the method – any changes to the parameter will not change the original value.

    pass by reference means that the address of the object is sent to the method – any changes to the parameter’s instance members will result in that change being made in the original value..

    What you are talking about in this article has nothing to do with pass by reference, instead you are talking about changing the address that the parameter points to, to some new location in memory — the address is constant but the data that the pointer refers to is not.

    1. miticosul says:

      +1! 🙂 well done! my exactly the same thoughts regarding this article!

      1. Pankaj says:

        Sorry to say this but your understanding about “Pass by Reference” and “Pass by Value” is not correct. Below quote properly describes it:

        Passing by reference means the called functions' parameter will be the same as the callers' passed argument (not the value, but the identity - the variable itself). Pass by value means the called functions' parameter will be a copy of the callers' passed argument. The value will be the same, but the identity - the variable - is different.

        Also refer this StackOverflow Post:

        1. Anilal says:

          It is still confusing.

          As you said in the beginning of the post,

          “Pass by Reference: An alias or reference to the actual parameter is passed to the method, that’s why it’s called pass by reference”

          And in the code example, exactly same happens when the method “foo” is called. Since the function argument is a reference to the original object, we can modify the content of the original object in method “foo”. ( balloon.setColor(“Red”);)

          So my understanding is
          1. For Primitive types, parameters are passed by value.
          2. For Object types, object reference is passed by value.

          1. szp says:

            As far as I know, the term “pass by value/reference” is comes form c/c++.
            But java actual do not features related the c/c++ pointer concept.
            So I suppose the point why people are debating here is that Mr Pankaj mixed the c/c++feature”pass by value/reference” with java in the title of the post.

            But, even through the mixture is confusing for me, I still learned some useful details of the implementation of java programming language. Thanks.

      2. Mohan says:

        I think, there is a serious misunderstanding in the article, regarding pass by reference. The swap method is just swapping the argument references and not the actual values, hence it doesn’t have effect. And as far as the definition is concerned, whatever pankaj wrote is correct, however his understanding is not. In pass by reference, instead of passing a copy of the parameters, we pass the reference or the identifier which holds the address of the parameters. And swapping references doesn’t mean swapping values which is the case here. Here is a sample program from C,

        void swap(int* a, int* b){ //Swapping the argument references
        int* c = NULL;
        c = a;
        a = b;
        b = c;


        void swap2(int** a, int** b){ //Swapping the actual references
        int* c = NULL;
        c = *a;
        *a = *b;
        *b = c;


        int main(){

        int *a = NULL;
        int *b = NULL;

        a = (int*) malloc ( sizeof(int));
        b = (int*) malloc ( sizeof(int));

        *a = 1;
        *b = 2;
        printf(“\n%d,%d”, *a , *b);
        printf(“\n%d,%d”, *a , *b);
        printf(“\n%d,%d”, *a , *b);
        return 0;

    2. Govind Ostwal says:

      +1 to William’s answer.
      Your test is not about pass by value / reference. Please replace it will valid test to avoid further confusion.
      Java is pass by value for primitives and pass by reference for objects.
      Please correct your misleading tutorial.

      1. Alice says:


        //pass by value for primitives
        public static void main(String[] args) {
        Integer a = 13;
        Integer b = 179;
        System.out.println(“a1:” + a + ” b1:” + b);//a1:13 b1:179
        swap(a, b);
        System.out.println(“a2:” + a + ” b2:” + b);//a2:13 b2:179
        System.out.println(“b3:” + b);//b3:179
        public static void swap(Integer c, Integer d) {
        System.out.println(“c1:” + c + ” d1:” + d);//c1:13 d1:179
        Integer temp = c;
        c = d;
        d = temp;
        System.out.println(“c2:” + c + ” d2:” + d + ” temp:” + temp);//c2:179 d2:13 temp:13
        public static void modify(Integer e) {
        System.out.println(“e1:” + e);//e1:179
        e = 241;
        System.out.println(“e2:” + e);//e2:241

        //pass by reference for objects
        public static void main(String[] args) {
        Balloon a = new Balloon(“Red”);
        Balloon b = new Balloon(“Blue”);
        System.out.println(“a1:” + a + ” ” + a.getColor());// Balloon@7852e922 Red
        System.out.println(“b1:” + b + ” ” + b.getColor());// Balloon@4e25154f Blue
        swap(a, b);
        System.out.println(“a2:” + a + ” ” + a.getColor());// Balloon@7852e922 Red
        System.out.println(“b2:” + b + ” ” + b.getColor());// Balloon@4e25154f Blue
        System.out.println(“b3:” + b + ” ” + b.getColor());// Balloon@4e25154f Red
        public static void swap(Balloon c, Balloon d) {
        System.out.println(“c1:” + c + ” ” + c.getColor());// Balloon@7852e922 Red
        System.out.println(“d1:” + d + ” ” + d.getColor());// Balloon@4e25154f Blue
        Balloon temp = c;
        c = d;
        d = temp;
        System.out.println(“c2:” + c + ” ” + c.getColor());// Balloon@4e25154f Blue
        System.out.println(“d2:” + d + ” ” + d.getColor());// Balloon@7852e922 Red
        System.out.println(“temp:” + temp + ” ” + temp.getColor());// Balloon@7852e922 Red
        public static void modify(Balloon e) {
        System.out.println(“e1:” + e + ” ” + e.getColor());// Balloon@4e25154f Blue
        System.out.println(“e2:” + e + ” ” + e.getColor());// Balloon@4e25154f Red
        e = new Balloon(“Green”);
        System.out.println(“e3:” + e + ” ” + e.getColor());// Balloon@70dea4e Green
        System.out.println(“e4:” + e + ” ” + e.getColor());// Balloon@70dea4e Blue

      2. Binith says:

        Agree that this is a misleading article, especially for new developers. What he says is correct, but its not the most relevant thing to understand. The important thing to understand is that Java uses pass by value for primitives and pass by reference for objects. The unfortunate thing is this article pops up as the first result in google and may people are thinking this is the right way. I would request to correct/rewrite it

        1. Pankaj says:

          Java is Pass by value, please read some more and correct your understanding. I also made a video, please watch that to understand this concept.

  43. Balwinder Singh says:

    Awesome explanation, thanks

  44. Pooya says:

    it is the best explenation about pass by reference and pass by value.
    everything is too clear right now

  45. Christopher says:


    I’m not a new programmer, but fairly new to Java. Once again your calm and helpful manner has come through for me. Thanks for the clarity.

    1. Pankaj says:

      Thanks for the kind words, appreciate it.

  46. this post clear my all problem.
    thank you
    keep posting

  47. Rahul says:

    Excellent explanation.

  48. Dharmendra says:

    Hi Pankaj,

    Thanks for such a wonderful article.
    I have one question based on your article.

    import java.util.ArrayList;
    import java.util.List;

    public class Test {

    public static void main(String[] args) {
    List a = new ArrayList();
    System.out.println(“LIST SIZE”+a.size());
    addInList(a, “Banana”);
    addInList(a, “COFFEE”);
    System.out.println(“LIST SIZE”+a.size());

    public static void addInList(List list, Object ojb){
    System.out.println(“LIST SIZE::”+list.size());


    LIST SIZE::2
    LIST SIZE::3

    If java support pass by value then why list size is modified
    In last print statement list size should be 1,

    Could you clarify to me


    1. Victor says:

      List is an object. When you pass a object to a method, you are passing the value of the memory address of the object to the method. Thus you can modify the attributes of the object. However, you cannot reassign another object to this memory address.

      1. Disha says:

        Hi Victor,

        In Java, the new keyword returns a reference to a newly-constructed object. Therefore, in

        String s = new String(“foo”);
        the value of variable s is a reference to a String object
        In the same way
        List a = new ArrayList();
        a should be reference to ArrayList Object

        Isn’t it?

  49. Bhagwati Lal says:

    Very Urgent please help me.

    Can you explain me how to draw memory diagram for following java code.

    public class ArrayOfReferences {

    public static void main(String[] args) {
    String[] names; /* How many objects do we have? */
    int numberOfNames = 3;

    Scanner scanner = new Scanner(;

    /* Reading names */
    names = new String[numberOfNames]; /* How many objects do we have? */

    for (int idx = 0; idx < names.length; idx++) {
    System.out.println("Enter the name of a friend: ");
    names[idx] =;

    /* Printing names */
    System.out.println("Your friends are: ");
    for (int idx = 0; idx < names.length; idx++) {


  50. Arunkumar says:

    Thanks a lot pankaj. That was a very good explanation.

  51. Jagadish says:

    Thanks Pankaj for explaining it more clearly in video. Now doubt is cleared.

  52. Vyacheslav says:

    Hello. Thanks for a good explanation.
    Yet, I suppose, there is an issue what to mean passing by value or by reference. Look there – I like very much the idea, that the main difference is:
    By val — The procedure code does not have any access to the underlying element in the calling code;
    By ref — The procedure code gives a direct reference to the programming element in the calling code.
    So, to my mind it is not clear that swap method is a correct way to obviously say if a language passes vars by value or by ref.

    1. Vyacheslav says:

      Well, also a good article: Java doesn’t support variable references, so, we can not speak at all about passing by reference in Java.

  53. Renat says:

    Man, you’re so confused it’s not funny. Stop writing stuff on the internet, people here believe all they see. Consider this – it’s an equivalent C program that passes a struct by reference and the behaviour is exactly the same as in Java:


    typedef struct s {
    const char *colour;
    } balloon;

    void swap (balloon *b1, balloon *b2) {
    balloon *tmp = b1;
    b1 = b2;
    b2 = tmp;

    int main (int argc, char **argv) {
    balloon b1, b2;
    b1.colour = “red”;
    b2.colour = “blue”;

    swap (&b1, &b2);

    printf (“red: %s, blue: %s\n”, b1.colour, b2.colour);

    return 0;

    1. Pankaj says:

      First of all here we are talking about Java. Even after reading and hopefully understanding above program, you don’t understood then I feel for you. Read again, may be somewhere else… nothing else I can do for you.

    2. Vlad says:

      Your code does not pass by reference, it passes pointer by value. Moreover, there is no such thing as “reference” in C, it exists only in C++:


      typedef struct s {
      const char *colour;
      } balloon;

      void swap (balloon &b1_r, balloon &b2_r) {
      balloon tmp = b1_r;
      b1_r = b2_r;
      b2_r = tmp;

      int main (int argc, char **argv) {
      balloon b1, b2;
      b1.colour = “red”;
      b2.colour = “blue”;

      swap (b1, b2);

      printf (“red: %s, blue: %s\n”, b1.colour, b2.colour);

      return 0;

  54. Praveen says:

    Damn it., That’s one of the simplest way to explain it.
    Thanks a ton Guru 🙂

  55. Gadrov says:

    I understand that Java is always pass by value, what I don’t understand is what exactly is the difference between pass an object by reference and pass an object reference by value.
    If you pass the reference value, isn’t that essentially the same thing? What would change in the swap and balloon examples?
    Currently I only understand the fact that calling the class’s constructor allocates a different memory set for the variable, otherwise I can’t see the difference. That’s why a lot of people say that Java passes objects by reference, because the behavior, if not identical, is extremely similar.

  56. Mandeep Sharma says:


    Can you please declare an array of integer and initialize it with some values. then pass this array to some method (return type void) and add 1 to all of the elements in array. and come back in main and try to print array element.. this will print the incremented values. Can you do clear this thing why its happens.

  57. Albert says:

    Dear Mr. Pankaj,

    I have a confusion with this method:

    private static void foo(Balloon balloon) { //baloon=100 // line1
    balloon.setColor(“Red”); //baloon=100
    balloon = new Balloon(“Green”); //baloon=200
    balloon.setColor(“Blue”); //baloon = 200
    In Line1,Could you please tell me why balloon points to “Blue” Object instead of “Red” Object ?

    1. Lakshman says:

      At line 1, object refers to Blue color Balloon object. From caller we passed Blue color Balloon object.

  58. Narendra Koli says:

    Thank you so much for creating video.
    Your tutorials are really helpful.

    1. Pankaj says:

      You are welcome Narendra.

  59. qiangyu says:

    good post ,Thanks

  60. Esakki says:

    Really good explanation (especially on the video content)

  61. Olga Guskova says:

    Very useful explanation and tutorial in whole.

  62. bin says:

    good one. Aprreciated

  63. Brihati Jain says:

    It was a great post. In video you are using stack memory for static methods memory allocation.Can you please explain why so?.Static methods exist in pergen space of heap

  64. Rahul Jain says:

    Very nicely explained

  65. Stefan Banu says:

    Hi Pankaj,

    from the video i understood exactly, so the video solution is much much better,

    thanks for all your work, your site is always a “reference” for my java development understanding.

    thanks and keep doing your thing.

    Stefan from Romania.

  66. Igorovics says:

    Great explanation, thank you.

  67. konrad says:

    I think that the post topic is somehow misleading/confusing. Of course Java is pass by value for primitives and pass by reference for objects – you even proved that drawing arrows (references!, not values).

    On the other hand those references (i.e. memory location addresses) are passed by value – that is how I think it should be understood.

    1. Bejond says:

      Totally agree with your opinion.

      I believe “Pass by Value: The method parameter values are copied to another variable and then the copied object is passed, that’s why it’s called pass by value.” is misunderstanding. Because right description should be “Pass by Value: The method parameter values are copied to another variable and then the copied reference is passed, that’s why it’s called pass by copied reference.”


  68. Pankaj Tandakar says:

    But in case of pass by value ,since a copy is passed the original one doesnt get changed which occurs in case of primitive data types ,but same we do in case of Objects ,its original value gets changed

  69. Dan says:

    These definitions are very counter intuitive. When your argument to a function is an object, java creates a copy of the “object reference” and passes that into the function scope. So, you are feeding in a reference to an object, which is a memory address, and because it is a copy of the original, Java becomes “pass by value.”

    Intuitively you would assume that passing in a reference somehow cannot be considered “pass by value” since it is called a reference.

    Definitions like this are part of the reason so many people are intimidated by coding.

    In my computer science courses, professors have started saying that objects as arguments in java are “pass by reference” because you are passing in their reference. Only way to change the status quo I suppose.

    1. Pankaj says:

      I loved your comment, this is the reason for all the confusion around it.

  70. Venkata Sriram says:

    Nice Explanation sir.Thanks

  71. Singh says:

    Thanks i was confused from last month..

  72. johny says:

    Hi Pankaj, thanks for including this nice post with good explanation. This post is very simple and easy to understand the reason for ‘why Java is pass by value but not by reference’. From your post it is clear that reference means reference to the memory location of the object. Each memory location holds a number i.e VALUE. This value is useful in identifying that memory location. So If there is a reference to an object, it means it is reference to the address location of that object. So, java refers the objects by their memory address. The name used to refer that memory location (i.e value) is termed as reference variable. As we are passing the reference of the memory location, java is pass by value (of the memory address of the object) but not pass by reference.
    One suggestion, if you would have added the explanation about primitive data types also, it would have helped many readers in understanding why swap method works in case of primitives but not objects.

    However, this is a very nice post right to the point.

  73. Manu M says:

    Before reading this post, i have understood this core concept, but realised it after i read this post.
    Java is always pass by value, value will be always primitive. Value can never be Object type, infact if we are passing object it seems we are passing the memory address of that object. This memory address is again an int value.
    Most of the books, authors failed to explain this. So here it is GREAT.

    “Java is pass by value, and sometimes pass through reference”.

  74. Victor says:

    So, java pass object by reference

  75. Victor says:

    Correctly to say that each object variable is reference and we can only pass this reference to method or reassign variable with new reference.

    I think you compare with C/C++. Where on language level is supported passing reference to reference (pointer to pointer). And data of object is not related to variable. We can pass address of variable which contain address where object is located. I think you understand me.

    In Java such functionality is not supported on language level. and to do this it require to create a new instance of your class with field which will hold a variable with reference to your object.

    For example, you want to pass Integer to method and change it in the body of that method but there is no method to change value of Integer object. So you need to create wrapper class for that Integer object and pass it to method.

  76. Praveen says:


    public class Test {

    public static void main(String[] args) {

    Balloon blue = new Balloon(“Blue”);

    private static void foo(Balloon balloon) {

    The output of the above program is Red.

    So, the above isn’t pass by reference ?

    1. Pankaj says:

      Think of it like two strings attached to a single balloon, the original one “blue” and at the time of invoking foo(), another one named “balloon”. Now if you will pull any of these, the balloon will be pulled towards you but as soon as you attached the “balloon” string to another balloon object, any further changes will not be done to the original object.

      This is “Pass by Value” and we are passing COPY of the reference.

      1. Praveen says:

        I am still confused. It seems java supports both pass by value and reference.

        In swap case it is pass by value. foo method in my previous comment acts as reference.

      2. Anonymous says:

        You got it all wrong man.

        You used strings in the above example, try using StringBuilder or StringBuffer, and see what you get.

        1. Pankaj says:

          I don’t know what you meant, can you comment with sample code?

    2. Abhi says:

      Hope below code might explain why its pass by value 🙂

      public static void main(final String[] args) {

      String test = “Check me”;

      public static void foo1(String str) {
      str = null;

      public static void foo2(String str) {
      str = str + “abhi”;


      Check me
      Check me

      1. James says:

        Apologies for late reply (6 years!) but String is a primitive hence you get those results. Java is pass-by-value for primitives and pass-by-reference for non-primitives.

  77. Atul says:

    So, if it is not by reference, so can you please explain the program as below :

    public class Test {

    public static void main(String[] args) {
    Employee emp1 = new Employee(“Employee 1”);

    System.out.println(“Call for Emp1 ” + emp1);


    System.out.println(“Name change DONE”);

    System.out.println(“Call for Emp1 ” + emp1);


    public static void nameChange(Employee e1){
    e1.setName(“Employee 3”);


    class Employee {
    String name;

    Employee(String nam){ = nam;

    public String getName() {
    return name;

    public void setName(String name) { = name;

    public String toString() {

    return “Object for “+;

    Becouse it is producing the output as

    Call for Emp1 Object for Employee 1
    Name change DONE
    Call for Emp1 Object for Employee 3

    1. Pankaj says:

      Again I would suggest read the full post, especially foo() method explanation for this.

  78. vijay sakhare says:

    This is vijay here u are having a very good interview questions which are very helpfull during interview.
    i have been asked one qus in gtt is that : purpose of class inside interface with example.
    i saw lot of blogs but doesnt get clearity…so plz post the answer for this qus also.

    Thx in advance.

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