Pass by reference versus value in variable versus array element

Question

I know this question apparently has many duplicates like here and here.

My question is different though.

Consider the following example:

public class MyClass {
    public static void test(int a, int b) {
        System.out.println("In test() at start: "+a+" "+b);
        int temp=a;
        a=b;
        b=temp;
        System.out.println("In test() at end: "+a+" "+b);
    }

    public static void main(String args[]) {
        int a=1, b=2;
        System.out.println("a: "+a+" b: "+b);

        test(a, b);
        System.out.println("a: "+a+" b: "+b);
    }
}

The output that I get for the above snippet is:

a: 1 b: 2
In test() at start: 1 2
In test() at end: 2 1
a: 1 b: 2

This shows that the original values of a and b in main() have not been swapped when I called test(), thereby implying (if I understand correctly) that it was passed by value.

Now, consider the following code snippet:

public class MyClass {
    public static void test(int[] arr) {
        System.out.println(arr[2]);
        arr[2]=20;
        System.out.println(arr[2]);
    }

    public static void main(String args[]) {
        int[] arr={0,1,2,3,4,5};
        System.out.println(arr[2]);

        test(arr);
        System.out.println(arr[2]);
    }
}

The output that I get for this code snippet is:

2
2
20
20

This shows that the value at arr[2] was changed in the original array in main(), thereby denoting (if I understand correctly) that the array was passed by reference.

Could someone please point out what is going on? Why does it show different behaviors?

Thanks!

All arrays are passed be reference, by default variables are passed by value. If you were to use `Integer` it would behave similarly — Mitch, May 25 '18 at 02:15
I come from a C++ background. So, if some one explains from _that perspective_ it would be very helpful for me. — , May 25 '18 at 02:15
@Mitchel0022, okay, got your point. `Integer` as the datatype of a variable or an array element, you mean? — , May 25 '18 at 02:25
Either, `Integer` is a object, therefore it will be passed by reference and changing it in the function will change it permanently. `int` will only change in the scope of the function — Mitch, May 25 '18 at 02:26
@Mitchel0022, okay. So to conclude, arrays and objects are passed by reference; everything else by value. What about containers? I think reference again (because they contain nothing but objects)? — , May 25 '18 at 02:27
Thank you, @Mitchel0022. It is very helpful for someone who comes from a C++ background like me. If you convert it into an answer, I would be glad to accept it. — , May 25 '18 at 02:29
Its not large enough of a question to justify an answer, but glad I could help — Mitch, May 25 '18 at 02:30
@StaticBeagle, guess what, the links that I have included at the beginning of the question is the exact same that you mention (maybe you didn't check those). And, my _entire post_ explains how it is different than that link. Do you still think that it is a duplicate?! — , May 25 '18 at 04:01
@RakeshKarandikar yes it is a duplicate and the accepted answer is not entirely correct. Everything in java is pass by value as pointed out in the answer in the link I provided. Read the answer in the old question carefully and you will realize that it is indeed a duplicate. — StaticBeagle, May 25 '18 at 04:16
@StaticBeagle, could you please elaborate? The opening paragraph itself says _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 is talking about objects. As per Mitchel0022's comments above, this is right; but variables of primitive data types are passed by value. Could you please confirm? — , May 25 '18 at 14:29
@RakeshKarandikar it creates a copy of the reference and passes that copy by value. Say you have `int[] array = ...` and you pass this array to a function `func(int[] arrayPrime)`, now inside the function `arrayPrime` points to the same location as `array`, thus `arrayPrime[2]` alters the value stored in `array[2]` but what happens if I do `arrayPrime = new int[]` inside the function? Now `arrayPrime` is no longer a reference to `array`. Think of a function passing a pointer `(int* p)`, C++ will pass a copy of than pointer to the function, thus passes the pointer by value. Clear as mud? =] — StaticBeagle, May 25 '18 at 21:55

Shanu Gupta · Answer 1 · 2018-05-25T03:07:42.473

I think you are just getting confused with the array part.

Array in java is a special object that contains references to other objects. You can very well change those "child" object's values as with any Object.

Remember this thumb rule that I keep to avoid confusion:

It doesn't matter what you do directly with the object but what you do inside it

Meaning that if you do some changes inside of any sub-object (or array item in your case) you'd see the side effects of it in your caller.

score -1 · Accepted Answer · answered May 25 '18 at 03:08

-1

Primitive datatype variables will be passed by value. Object datatype variables will be passed by reference. Array is Object datatype, so it will be passed by reference.

answered May 25 '18 at 03:08

user1746345

14

This is incorrect. Please read the canonical duplicate link. It's a subtle but _very important_ distinction, because in Java there are only primitives and references. Everything is passed by value. – Jim Garrison May 25 '18 at 04:18
@JimGarrison, what about the array then? As Mitchell0022 pointed out in the comments above, I think they are passed by reference, because that explains the output and observation of the second code snippet. Could you please elaborate to help me understand? – May 25 '18 at 14:30

Pass by reference versus value in variable versus array element

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