Add an object to an ArrayList and modify it later

Question

If I have an ArrayList, and I added an object to it, and later I modified this object, will this change reflect in the ArrayList? or when I add the object to the ArrayList, Java creates a copy and add it to the ArrayList?

What if I change the reference to this object to null? Does that mean that the object in the ArrayList now null too?

This question (& answer) really deserves a bookmark! – Terry Oct 30 '13 at 13:47 — Terry, Oct 30 '13 at 13:47
Just all questions I was about to ask:) – navid Dec 14 '18 at 08:49 — navid, Dec 14 '18 at 08:49

score 90 · Accepted Answer · edited Jun 20 '20 at 09:12

90

will this change reflect in the ArrayList?

Yes, since you added a reference to the object in the list. The reference you added will still point to the same object, (which you modified).

or when I add the object to the ArrayList, Java creates a copy and add it to the ArrayList?

No, it won't copy the object. (It will copy the reference to the object.)

What if I change the reference to this object to null? Does that mean that the object in the ArrayList now null too?

No, since the content of the original reference was copied when added to the list. (Keep in mind that it is the reference that is copied, not the object.)

Demonstration:

StringBuffer sb = new StringBuffer("foo");

List<StringBuffer> list = new ArrayList<StringBuffer>();
list.add(sb);

System.out.println(list);   // prints [foo]
sb.append("bar");

System.out.println(list);   // prints [foobar]

sb = null;

System.out.println(list);   // still prints [foobar]

edited Jun 20 '20 at 09:12

Community

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answered Aug 16 '11 at 15:14

aioobe

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what about this: Button a = new Button; Button b = new Button; Button current = a; List.add(current); current = b; If i print the list content it is gonna be "a" or "b"? – Lorenzo Sciuto Sep 10 '13 at 07:23
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@aioobe Your answer seems to conflict with the first point in your post where you state that the reference to the object is copied (in this case, 'current'), which is then changed from a to b. I would've expected that it prints b. Can you elaborate? – Jan K. Oct 19 '13 at 15:28
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Just keep in mind that Java is always [pass by value](http://stackoverflow.com/questions/373419/whats-the-difference-between-passing-by-reference-vs-passing-by-value) (in particular references to objects are passed by value) and it should all be clear. – aioobe Oct 19 '13 at 17:12
this doesn't seem to be the case for adding a `String` and then changing it. i.e. `String a = "first"; list.add(a); a = "second"; ...print(list.get(0)) // "first"` – Don Cheadle Mar 06 '15 at 17:09
I think you may want to clarify what is meant by `pass by value` as that would seem to imply that you what was sent cannot be altered as the value has been sent, not a reference to something – Don Cheadle Mar 06 '15 at 17:28
@mmcrae, You sound confused. *[...] the case for adding a `String` and then changing it [...]* You can't change Strings. They are immutable. – aioobe Mar 06 '15 at 18:33

score 7 · Answer 2 · answered Aug 16 '11 at 15:16

7

Any change to the object will be reflected in the list.

However, when you deal with objects like Strings that are immutable, a new object will be created on "change operations". Than actually your old object is still in the list while you got a new one elsewhere.

answered Aug 16 '11 at 15:16

DerMike

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this is a very important distinction – Don Cheadle Mar 06 '15 at 17:20

score 1 · Answer 3 · answered Dec 14 '18 at 09:12

Thank you all guys. I figured it out again by reading your posts. It is probably a confusing concept, since I've had it digested long ago but recently I forgot how it works again. So I wanna share the key that solves the problem for me. (note that non-primitive Objects (primitives are int, boolean, etc) in java are technically pointers). When u add the object o to a list, the list and the o points to the same thing so when u modify o the list's item changes. This is as long as they point to the same thing and breaks when o is pointing to something else by =.

o = null;   //o points to nothing and changes in o from now on doesn't effect the list's item

or

Object a = new Object();
o = a;    //o and the list's item don't point to same thing so changes in o doesn't effect the list's item (but it effects a)

hope it helps someone

score -1 · Answer 4 · answered Jan 15 '15 at 17:45

Wanted to add another demonstration where the ArrayList is inside of a Map as the value. The ArrayList is modified after adding to the Map and the Map reflects the changes.

The Map has one element with mother's name as the key and children as the value.

    String key = "adeleMom";
    Map<String, ArrayList<String>> myMap = new HashMap<String, ArrayList<String>>();
    ArrayList<String> firstList = new ArrayList<String>();
    firstList.add("adele");
    myMap.put(key, firstList);
    firstList = null;
    ArrayList secondList = myMap.get(key);
    System.out.println(secondList); // prints [adele]
    secondList.add("bonnie");
    System.out.println("Added bonnie");
    ArrayList thirdList = myMap.get(key);
    System.out.println(thirdList); // prints [adele, bonnie]

Add an object to an ArrayList and modify it later

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