Java Arraylist size

Question

I'm having an issue understanding the output of size() for a "nested" ArrayList:

ArrayList<ArrayList<ArrayList<ArrayList<String>>>> myEmpls = new ArrayList<>();
ArrayList<ArrayList<ArrayList<String>>> dummy2 = new ArrayList<>();
ArrayList<ArrayList<String>> dummy = new ArrayList<>();

        for (int x = 0; x < 4; x++) {
        myEmpls.add(dummy2);
    }

    System.out.println(myEmpls.size());

    // returns 4 as expected

    System.out.println(myEmpls.get(0).size());
    System.out.println(myEmpls.get(1).size());

    // both return 0 as expected

    for (int x = 0; x < 10; x++) {
        myEmpls.get(0).add(dummy);
    }
    System.out.println(myEmpls.get(1).size());

    // returns 10 altough I added to myEmpls.get(0) and not 
    // myEmpls.get(1) so I expected this to still be 0.

Can someone explain why this is the case? Maybe I am missing something obvious.

Thanks in advance.

Thanks for all the answers, I actually ran into the same issue before but since I havent been coding for some time I failed to recognize it! — user2299050
– user2299050, Commented May 3, 2015 at 16:19

libik · Accepted Answer · 2015-05-03 16:12:31Z

3

You are adding same reference of dummy2 to all four positions of myEmpls. Therefore when you access any of them, you access same object as having variable dummy2, which really has 10 objects in it, because you added it in it.

You can even do this and it also returns 10

System.out.println(dummy2.size());

If you do this

for (int x = 0; x < 4; x++) {
    myEmpls.add(new ArrayList<ArrayList<ArrayList<String>>>());
}

It would behave as you expected

answered May 3, 2015 at 16:12

libik

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Comments

SamTebbs33 · Accepted Answer · 2015-05-03 16:13:46Z

1

The reason is that you have created an ArrayList object (dummy2) and then added that exact object to the list 4 times. Therefore, when you add to dummy to the object from myEmpls.get(0), you are adding dummy to the other 3 indexes too as they are all the same object. Hence, the size is 10. Each index inside myEmpls points to exactly the same address in memory and therefore whatever you do to one of the indexes, you do to all of the others.

answered May 3, 2015 at 16:13

SamTebbs33

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Akash Thakare · Accepted Answer · 2015-05-03 16:14:55Z

1

List keeps the reference of Object added in it,change in state of Object reflects to all the references added in the list in your case.

ForExample :

    ArrayList<List<String>> mainList =new ArrayList<>();
    List<String> test1 =  new ArrayList<>();
    mainList.add(test1);
    mainList.add(test1);
    mainList.add(test1);
    mainList.add(test1);

    System.out.println(mainList);

    test1.add("Hello");//now add item to test1

    System.out.println(mainList);

OUTPUT :

[[], [], [], []]
[[Hello], [Hello], [Hello], [Hello]]

answered May 3, 2015 at 16:14

Akash Thakare

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Eran · Accepted Answer · 2015-05-03 16:18:20Z

1

myEmpls.add(dummy2); add the same ArrayList instance to myEmpls muptiple times.

Therefore, calling myEmpls.get(0).add(dummy) adds elements to the same ArrayList instance referred by myEmpls.get(1). Therefore the size() of myEmpls.get(1) is 10, since myEmpls(0)==myEmpls(1).

If you change your first loop to :

for (int x = 0; x < 4; x++) {
    dummy2 = new ArrayList<>();
    myEmpls.add(dummy2);
}

each element of myEmpls will be distinct, and myEmpls.get(1).size() would return 0 as you expected.

edited May 3, 2015 at 16:18

answered May 3, 2015 at 16:12

Eran

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