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Is Java "pass-by-reference" or "pass-by-value"?


Question

I always thought Java uses pass-by-reference.

However, I've seen a couple of blog posts (for example, this blog) that claim that it isn't (the blog post says that Java uses pass-by-value).

I don't think I understand the distinction they're making.

What is the explanation?

2020/08/06
1
6700
8/6/2020 7:24:06 AM


I just noticed you referenced my article.

The Java Spec says that everything in Java is pass-by-value. There is no such thing as "pass-by-reference" in Java.

The key to understanding this is that something like

Dog myDog;

is not a Dog; it's actually a pointer to a Dog.

What that means, is when you have

Dog myDog = new Dog("Rover");
foo(myDog);

you're essentially passing the address of the created Dog object to the foo method.

(I say essentially because Java pointers aren't direct addresses, but it's easiest to think of them that way)

Suppose the Dog object resides at memory address 42. This means we pass 42 to the method.

if the Method were defined as

public void foo(Dog someDog) {
    someDog.setName("Max");     // AAA
    someDog = new Dog("Fifi");  // BBB
    someDog.setName("Rowlf");   // CCC
}

let's look at what's happening.

  • the parameter someDog is set to the value 42
  • at line "AAA"
    • someDog is followed to the Dog it points to (the Dog object at address 42)
    • that Dog (the one at address 42) is asked to change his name to Max
  • at line "BBB"
    • a new Dog is created. Let's say he's at address 74
    • we assign the parameter someDog to 74
  • at line "CCC"
    • someDog is followed to the Dog it points to (the Dog object at address 74)
    • that Dog (the one at address 74) is asked to change his name to Rowlf
  • then, we return

Now let's think about what happens outside the method:

Did myDog change?

There's the key.

Keeping in mind that myDog is a pointer, and not an actual Dog, the answer is NO. myDog still has the value 42; it's still pointing to the original Dog (but note that because of line "AAA", its name is now "Max" - still the same Dog; myDog's value has not changed.)

It's perfectly valid to follow an address and change what's at the end of it; that does not change the variable, however.

Java works exactly like C. You can assign a pointer, pass the pointer to a method, follow the pointer in the method and change the data that was pointed to. However, you cannot change where that pointer points.

In C++, Ada, Pascal and other languages that support pass-by-reference, you can actually change the variable that was passed.

If Java had pass-by-reference semantics, the foo method we defined above would have changed where myDog was pointing when it assigned someDog on line BBB.

Think of reference parameters as being aliases for the variable passed in. When that alias is assigned, so is the variable that was passed in.

2015/08/28

Java always passes arguments by value, NOT by reference.


Let me explain this through an example:

public class Main {

     public static void main(String[] args) {
          Foo f = new Foo("f");
          changeReference(f); // It won't change the reference!
          modifyReference(f); // It will modify the object that the reference variable "f" refers to!
     }

     public static void changeReference(Foo a) {
          Foo b = new Foo("b");
          a = b;
     }

     public static void modifyReference(Foo c) {
          c.setAttribute("c");
     }

}

I will explain this in steps:

  1. Declaring a reference named f of type Foo and assign it a new object of type Foo with an attribute "f".

     Foo f = new Foo("f");
    

    enter image description here

  2. From the method side, a reference of type Foo with a name a is declared and it's initially assigned null.

     public static void changeReference(Foo a)
    

    enter image description here

  3. As you call the method changeReference, the reference a will be assigned the object which is passed as an argument.

     changeReference(f);
    

    enter image description here

  4. Declaring a reference named b of type Foo and assign it a new object of type Foo with an attribute "b".

     Foo b = new Foo("b");
    

    enter image description here

  5. a = b makes a new assignment to the reference a, not f, of the object whose attribute is "b".

    enter image description here

  6. As you call modifyReference(Foo c) method, a reference c is created and assigned the object with attribute "f".

    enter image description here

  7. c.setAttribute("c"); will change the attribute of the object that reference c points to it, and it's the same object that reference f points to it.

    enter image description here

I hope you understand now how passing objects as arguments works in Java :)

2020/08/29

This will give you some insights of how Java really works to the point that in your next discussion about Java passing by reference or passing by value you'll just smile :-)

Step one please erase from your mind that word that starts with 'p' "_ _ _ _ _ _ _", especially if you come from other programming languages. Java and 'p' cannot be written in the same book, forum, or even txt.

Step two remember that when you pass an Object into a method you're passing the Object reference and not the Object itself.

  • Student: Master, does this mean that Java is pass-by-reference?
  • Master: Grasshopper, No.

Now think of what an Object's reference/variable does/is:

  1. A variable holds the bits that tell the JVM how to get to the referenced Object in memory (Heap).
  2. When passing arguments to a method you ARE NOT passing the reference variable, but a copy of the bits in the reference variable. Something like this: 3bad086a. 3bad086a represents a way to get to the passed object.
  3. So you're just passing 3bad086a that it's the value of the reference.
  4. You're passing the value of the reference and not the reference itself (and not the object).
  5. This value is actually COPIED and given to the method.

In the following (please don't try to compile/execute this...):

1. Person person;
2. person = new Person("Tom");
3. changeName(person);
4.
5. //I didn't use Person person below as an argument to be nice
6. static void changeName(Person anotherReferenceToTheSamePersonObject) {
7.     anotherReferenceToTheSamePersonObject.setName("Jerry");
8. }

What happens?

  • The variable person is created in line #1 and it's null at the beginning.
  • A new Person Object is created in line #2, stored in memory, and the variable person is given the reference to the Person object. That is, its address. Let's say 3bad086a.
  • The variable person holding the address of the Object is passed to the function in line #3.
  • In line #4 you can listen to the sound of silence
  • Check the comment on line #5
  • A method local variable -anotherReferenceToTheSamePersonObject- is created and then comes the magic in line #6:
    • The variable/reference person is copied bit-by-bit and passed to anotherReferenceToTheSamePersonObject inside the function.
    • No new instances of Person are created.
    • Both "person" and "anotherReferenceToTheSamePersonObject" hold the same value of 3bad086a.
    • Don't try this but person==anotherReferenceToTheSamePersonObject would be true.
    • Both variables have IDENTICAL COPIES of the reference and they both refer to the same Person Object, the SAME Object on the Heap and NOT A COPY.

A picture is worth a thousand words:

Pass by Value

Note that the anotherReferenceToTheSamePersonObject arrows is directed towards the Object and not towards the variable person!

If you didn't get it then just trust me and remember that it's better to say that Java is pass by value. Well, pass by reference value. Oh well, even better is pass-by-copy-of-the-variable-value! ;)

Now feel free to hate me but note that given this there is no difference between passing primitive data types and Objects when talking about method arguments.

You always pass a copy of the bits of the value of the reference!

  • If it's a primitive data type these bits will contain the value of the primitive data type itself.
  • If it's an Object the bits will contain the value of the address that tells the JVM how to get to the Object.

Java is pass-by-value because inside a method you can modify the referenced Object as much as you want but no matter how hard you try you'll never be able to modify the passed variable that will keep referencing (not p _ _ _ _ _ _ _) the same Object no matter what!


The changeName function above will never be able to modify the actual content (the bit values) of the passed reference. In other word changeName cannot make Person person refer to another Object.


Of course you can cut it short and just say that Java is pass-by-value!

2018/01/04

Java is always pass by value, with no exceptions, ever.

So how is it that anyone can be at all confused by this, and believe that Java is pass by reference, or think they have an example of Java acting as pass by reference? The key point is that Java never provides direct access to the values of objects themselves, in any circumstances. The only access to objects is through a reference to that object. Because Java objects are always accessed through a reference, rather than directly, it is common to talk about fields and variables and method arguments as being objects, when pedantically they are only references to objects. The confusion stems from this (strictly speaking, incorrect) change in nomenclature.

So, when calling a method

  • For primitive arguments (int, long, etc.), the pass by value is the actual value of the primitive (for example, 3).
  • For objects, the pass by value is the value of the reference to the object.

So if you have doSomething(foo) and public void doSomething(Foo foo) { .. } the two Foos have copied references that point to the same objects.

Naturally, passing by value a reference to an object looks very much like (and is indistinguishable in practice from) passing an object by reference.

2016/01/22

Java passes references by value.

So you can't change the reference that gets passed in.

2008/09/02

Source: https://stackoverflow.com/questions/40480
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