CODEWITHSUNDEEP
javareferencearraylistpass-by-reference
fvgs
fvgs

Reputation: 21982

Passing ArrayList as value only and not reference

Simply put, I have a method with an ArrayList parameter. In the method I modify the contents of the ArrayList for purposes relevant only to what is returned by the method. Therefore, I do not want the ArrayList which is being passed as the parameter to be affected at all (i.e. not passed as a reference).

Everything I have tried has failed to achieve the desired effect. What do I need to do so that I can make use of a copy of the ArrayList within the method only, but not have it change the actual variable?

Upvotes: 42

Views: 68295

Answers (8)

Parth
Parth

Reputation: 51

On the lines of the existing answers but using the ArrayList API. You can use subList(fromIndex, toIndex) method. It explicitly creates a view of the list with only desired elements (of course, in sequence). Here, even if you modify the view with add/remove etc operations, it won't change the original list. It saves you from explicitly creating a copy.

Something like this:

public void recursiveMethod(List<Integer> list) {
    if(base)
         return;
    recursiveCall(list);

    // following will just create a tail list but will not actually modify the list
    recursiveCall(list.subList(1, list.size());
}

Upvotes: 0

Avi
Avi

Reputation: 21858

Even if you had a way to pass the array list as a copy and not by reference it would have been only a shallow copy.

I would do something like:

void foo(final ArrayList list) {

    ArrayList listCopy = new ArrayList(list);
    // Rest of the code

}

And just work on the copied list.

Upvotes: 56

Caio Faustino
Caio Faustino

Reputation: 165

I'm not sure on why, even after new ArrayList<MyObj>(old) the object was still changing reference in places it wasn't supposed to. So I had to instantiate a new copy of the objects inside.

I made a copy constructor like the one on the ArrayList and did like

 newArray = new ArrayList<MyObj>();
        for (int i = 0; i < oldArray.size(); i++) {
            newArray.add(new MyObj(ondArray.get(i)));
        }

Just hope to help someone else if the answer from Avi is not enough in your case, like mine with a code too messy to even understand =P

Upvotes: 3

Ankur Shanbhag
Ankur Shanbhag

Reputation: 7804

Try this in you method :

void method(List<Integer> list) {
        List copyList =  new ArrayList<Integer>();
        copyList.addAll(list); // This will create a copy of all the emlements of your original list
    }

Upvotes: 3

bchociej
bchociej

Reputation: 1399

You can create a copy of the ArrayList using ArrayList's copy constructor:

ArrayList copy = new ArrayList(original);

But if the elements of the list are also objects, then you must be aware that modifying a member of the copy will also modify that member in the original.

Upvotes: 22

Sri Harsha Chilakapati
Sri Harsha Chilakapati

Reputation: 11950

Just clone it.

public ArrayList cloneArrayList(ArrayList lst){
    ArrayList list = new ArrayList();
    for (int i=0; i<lst.size(); i++){
        list.add(lst.get(i));
    }
    return list;
}

Add suggested in the comments, you can also use

ArrayList copy = new ArrayList(original);

and also

ArrayList copy = new ArrayList();
copy.addAll(original);

Upvotes: 1

Luiggi Mendoza
Luiggi Mendoza

Reputation: 85779

You could pass Collections#unmodifiableList(yourList) in order to send an unmodifiable copy of your list. By the way, your List<Whatever> is passed by value since Java always pass by value, note that in foo(List<Whatever> list) method you can not modify the list value but you can modify its contents.

public class MyClass {

    List<Whatever> list = new ArrayList<Whatever>();

    public void bar() {
        //filling list...
        foo(Collections.unmodifiableList(list));
    }

    public void foo(List<Whatever> list) {
        //do what you want with list except modifying it...
    }
}

Upvotes: 6

helion3
helion3

Reputation: 37381

You could use the .clone method or a CopyOnWriteArrayList to make a copy, thereby not impacting the original.

Upvotes: 3

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