ArrayList removes object from an other ArrayList?

Question

So, I have this use case:

ArrayList<String> test1 = new ArrayList<String>();
test1.add("string1");
test1.add("string2");

ArrayList<String> test2 = test1;
test2.remove(0);
System.out.println(test1.get(0)); // Expect to see "string1"

The first ArrayList test1, has two String elements. Then I make a new (?) ArrayList, test2, which is the same as test1. When I remove the first element from test2 (which is "string1") and then try to display the first element from test1, it returns "string2"... also there the first element "string1" is removed somehow.

How is that possible?

you're using the same instance of the arraylist in both variables — pedrohreis
– pedrohreis, Commented May 24, 2017 at 16:49

Suresh Atta · Accepted Answer · 2017-05-24 16:49:05Z

1

Then I make a new (?) ArrayList, test2, which is the same as test1.

No you haven't. You have just pointed your arraylist 2 to arraylist 1.

You can either use a Shallow constructor of list or use addAll method.

List<String> test2 = new ArrayList<String>();
test2.addAll(test1);

answered May 24, 2017 at 16:49

Suresh Atta

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dumbPotato21 · Accepted Answer · 2017-05-24 16:51:20Z

0

You are assigning the reference of test1 to test2. This means, that both the ArrayList refer to the same instance,

Therefore, methods called on one, will virtually reflect in the other one too, as they refer to the same object.

To achieve what you want, do

List<String> test2 = new ArrayList<>(test1);

This copies the contents of the test1 to test2.

answered May 24, 2017 at 16:51

dumbPotato21

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Pablo Rosales · Accepted Answer · 2017-05-24 17:05:39Z

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Create a new ArrayList with the same elements as test1

ArrayList<String> test2 = new ArrayList<>(test1)

edited May 24, 2017 at 17:05

answered May 24, 2017 at 16:52

Pablo Rosales

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1 Comment

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Why not just ArrayList<String> test2 = new ArrayList<>(test1)? What do you think Arrays.asList() adds to the picture?

Sadiq Ali · Accepted Answer · 2017-05-24 17:27:09Z

    ArrayList<String> test1 = new ArrayList<String>();
    test1.add("string1");
    test1.add("string2");

    ArrayList<String> test2 = test1;
    test2.remove(0);
    System.out.println(test1.get(0)); // Expect to see "string1"

    ArrayList<String> test3 = new ArrayList<>();

Upon running the code in debug you can see that the reference of test1 and test2 is pointing to the same location. That is because you are not copying the values but assigning the same reference to the other list. Notice that the reference of test3 is different than the other 2.

In order to just copy the values you can use

test2 = new ArrayList<Object>(test1);

Please find the image attached below to view the reference value.

Result of running the code stated above in debug

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