JAVA, list of lists

Question

I have list and list of lists:

ArrayList<String> singleList = new ArrayList<String>();
ArrayList<ArrayList<String>> listOfLists = new ArrayList<ArrayList<String>>();

I do not understand the behavior of these lists. I decided to show you a simple example:

listOfLists.clear();

singleList.clear();
singleList.add("A");
singleList.add("B");
singleList.add("C");
listOfLists.add(singleList);

singleList.clear();
singleList.add("D");
singleList.add("E");
singleList.add("F");
listOfLists.add(singleList);

singleList.clear();
singleList.add("G");
singleList.add("H");
singleList.add("I");
listOfLists.add(singleList);

for(int x = 0; x < listOfLists.size(); x++)
{
    for(int z = 0; z < singleList.size(); z++)
        {
        System.out.print(listOfLists.get(x).get(z));
        System.out.print(" ");
    }
    System.out.println("");
}

And the result I got was:

G H I G H I G H I

Instead:

A B C D E F G H I

Where is a problem with my thinking? What should I do to get result as above?

Because when you add singleList to listOfLists you do not make a copy of it, you add a reference to first list inside second one (so you won't even duplicate its content). — Adriano Repetti
– Adriano Repetti, Commented Oct 26, 2013 at 10:55

Guillaume Poussel · Accepted Answer · 2013-10-26 10:56:28Z

15

Objects are always passed as references in Java.

When you add singleList to listOfLists, you are in fact adding a reference to singleList. Since you've added it 3 times, you got the current value of singleList, repeated 3 times.

The "previous values" of singleList are stored nowhere, so A B C and D E F are lost.

You need to make a copy of your list, by using new ArrayList<String>(singleList). Then, add this copy to listOfLists.

answered Oct 26, 2013 at 10:56

Guillaume Poussel

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1 Comment

Mariusz Jucha Over a year ago

It works. Thanks. Also now I understand what was wrong with my thinking.

Evans · Accepted Answer · 2013-10-26 10:57:01Z

The problem is how object references work. Going step by step

singleList.clear();
singleList.add("A");
singleList.add("B");
singleList.add("C");
listOfLists.add(singleList);

//singleList -> A, B, C 
//listOfLists -> singleList

singleList.clear();
singleList.add("D");
singleList.add("E");
singleList.add("F");
listOfLists.add(singleList);

//singleList -> D, E, F
//listOfLists -> singleList, singleList

singleList.clear();
singleList.add("G");
singleList.add("H");
singleList.add("I");
listOfLists.add(singleList);

//singleList -> G, H, I
//listOfLists -> singleList, singleList, singleList

Now, you are printing listOfLists, wich contains 3 times singleList. But singleList contains now G, H, I

To get the desired result, you need to use different lists, one with A, B, C, other with D, E, F, and another one with G, H, I.

singleList1 -> A, B, C
singleList2 -> D, E, F
singleList3 -> F, G, H
listOfLists -> singleList1, singleList2, singleList3

user4857117 · Accepted Answer · 2015-05-02 13:52:22Z

0

Instead of having different variables for singleList, you can also do this when adding to listOfLists: listOfLists.add(new ArrayList(singleList));

this creates a copy of singlelist that has a different memory location unlike the one you did which refers to the same location.

answered May 2, 2015 at 13:52

user4857117

1

Comments

Personinous Apraham · Accepted Answer · 2015-09-05 13:02:08Z

0

To make a list of lists, you can easily use ArrayList<ArrayList<Integer>> listOfLists = new ArrayList<ArrayList<Integer>>; to make a list of lists of integers, or replaces Integer with String or whatever you want. To access items, use listOfLists.get(0).get(0) for example, to get the first item of the first list inside of listOfLists. Not sure if this helps.

answered Sep 5, 2015 at 13:02

Personinous Apraham

1

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