Is Java "pass

I always thought Java was pass-by-reference . However, I've seen a couple of blog posts (for example, this blog) that claim it isn't. I don't think I understand the distinction they're making.

What is the explanation?

Java is always pass-by-value . Unfortunately, they decided to call the location of an object a "reference". When we pass the value of an object, we are passing the reference to it. This is confusing to beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true, java passes by value
    aDog.getName().equals("Fifi"); // false 
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max" . The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo .

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...) . Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog itself (except when d is changed to point to a different Dog instance like d = new Dog("Boxer") ).

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.

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.

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:

Foo f = new Foo("f");

public static void changeReference(Foo a)

changeReference(f);

Foo b = new Foo("b");

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

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