Java array modifying itself due to pass-by-something error?

Question

I'm instantiating one array 'array1' by another 'array2' and then attempting to modify 'array2', which changes 'array1'. After many hours I realised that it may be a call-by- or a pass-by- ref/value error. Can someone please help me solve this and put me in the right direction?

    int[] src = {0,4,3,2,1};
    int[] dest = src;

    dest[0] = dest[0] + 2;

    for (int node: dest) {
        System.out.print(node + " ");
    }

    System.out.println("");
    for (int node: src) {
        System.out.print(node + " ");
    }

This produces:

2 4 3 2 1

i.e. the source array gets modified too. Thanks in advance.

Both objects (src and dest) hold the same reference, so any change in the state on one of them will be reflected in the other. — Luiggi Mendoza
– Luiggi Mendoza, Commented Feb 7, 2014 at 17:56
Create different arrays and initialize each entry with the same value. You can use System.arrayCopy and save some time. — Luiggi Mendoza
– Luiggi Mendoza, Commented Feb 7, 2014 at 17:57

Community · Accepted Answer · 2017-05-23 12:20:40Z

5

When you do

int[] dest = src;

both variables are referencing to the same position in memory. In other words, they are referencing to the same array. You can try using Arrays.copyOf:

int[] dest = Arrays.copyOf(src, src.length);

Edit: As @BackSlach commented, you can also use clone():

int[] dest = src.clone(); //clone() is a method present in every object in java

As a third alternative you can use System.arrayCopy()

int[] dest = new int[src.length];
System.arraycopy(src, 0, dest, 0, src.length);

Note that they have similar performance.

edited May 23, 2017 at 12:20

CommunityBot

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answered Feb 7, 2014 at 17:57

Christian Tapia

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6 Comments

BackSlash Over a year ago

No need for Arrays.copyOf, just use src.clone(), arrays are objects in java :)

Goldfishy Over a year ago

Thanks. So far Arrays.copyOf, .clone() and System.arrayCopy are being suggested. Which one is best?

arshajii Over a year ago

You should stay away from clone(). Java's cloning mechanism is largely broken.

BackSlash Over a year ago

@arshajii Not really true. It's recommended to stay away from that method for objects, but for arrays it's perfectly fine.

arshajii Over a year ago

@BackSlash You said yourself: "arrays are objects in java".

|

rgettman · Accepted Answer · 2014-02-07 17:58:08Z

2

You will need to create a copy of the src array in dest, so that when you modify dest, you don't also modify src. Just saying

int[] dest = src;

doesn't copy the array; it just creates another references to the same array. Create a copy:

int[] dest = new int[src.length];
System.arraycopy(src, 0, dest, 0, src.length);

answered Feb 7, 2014 at 17:58

rgettman

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Comments

Ben Griffiths · Accepted Answer · 2014-02-07 18:03:07Z

0

You are thinking on the right lines: everything that isn't a primitive type is passed by reference in Java.

When the following is executed, you are declaring a new reference to the same array in memory:

int[] dest = src;

So you need to declare a new array. You could then either copy the values of the elements of the first array to the second one 'manually' in a loop, or use the arraycopy method in the System namespace

http://docs.oracle.com/javase/7/docs/api/java/lang/System.html#arraycopy

answered Feb 7, 2014 at 18:03

Ben Griffiths

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Comments

arshajii · Accepted Answer · 2014-02-07 18:04:34Z

It has nothing to do with calling/passing. When you do

int[] dest = src;

then dest points to the same object that src points to. Therefore, a modification in dest can also be seen in src:

           +---+---+---+---+---+
src -----> | 0 | 4 | 3 | 2 | 1 | <----- dest
           +---+---+---+---+---+


dest[0] = dest[0] + 2;


           +---+---+---+---+---+
src -----> | 2 | 4 | 3 | 2 | 1 | <----- dest
           +---+---+---+---+---+

src[0] is 2 now as well, even though you modified dest.

As has been suggested, you can use Arrays.copyOf() to produce a new, distinct array with the same contents as src.

int[] dest = Arrays.copyOf(src, src.length);

Now, we have this situation:

            +---+---+---+---+---+
src  -----> | 0 | 4 | 3 | 2 | 1 |
            +---+---+---+---+---+

            +---+---+---+---+---+
dest -----> | 0 | 4 | 3 | 2 | 1 |
            +---+---+---+---+---+


dest[0] = dest[0] + 2;


            +---+---+---+---+---+
src  -----> | 0 | 4 | 3 | 2 | 1 |
            +---+---+---+---+---+

            +---+---+---+---+---+
dest -----> | 2 | 4 | 3 | 2 | 1 |
            +---+---+---+---+---+

Notice that the modification to dest did not affect src.

markusthoemmes · Accepted Answer · 2014-02-07 18:07:27Z

0

You're right, thats the source of your problem.

What you can do about it is to write a copy method that loops over your array and writes all the numbers to the new array.

Please keep in mind that this will only work for primitive types like int. If you have objects in your array, they are passed by reference aswell so deep-copying an array of any object needs some more work.

Also check out the clone() method, which can be overloaded to provide a mechanism to copy an object you're implementing.

edited Feb 7, 2014 at 18:07

answered Feb 7, 2014 at 17:57

markusthoemmes

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3 Comments

markusthoemmes Over a year ago

@Christian oh, wasn't aware of that. You never stop learning, thanks for mentioning :).

BackSlash Over a year ago

Also check out the clone() method, which can be overloaded to provide a mechanism to copy an object. Not in this case: you can't extend array, so you can't overload (override) that method.

markusthoemmes Over a year ago

@BackSlash Yes I'm aware of that, but i thought it would be worth mentioning it to make the OP aware of future problems with Object copying.

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