Problem with assigning an array to other array in Java

Question

public class TestingArray {

    public static void main(String[] args) {

        int iCheck = 10;
        int j = iCheck;
        j = 11;
        System.err.println("value of iCheck "+iCheck);

        int[] val1 = {1,2,9,4,5,6,7};
        int[] val2 = val1;
        val2[0] = 200;
        System.err.println("Array Value "+val1[0]);
    }
}

Output:

value of iCheck 10
Array Value 200

From the above code, I found that if any array val2 is being assigned to another array val1 and if we change any value of val2 array, the result is as well reflected for the array val1 while the same scenario is not with variable assignment. Why?

NPE · Accepted Answer · 2011-09-08 15:13:21Z

39

The following statement makes val2 refer to the same array as val1:

int[] val2 = val1;

If you want to make a copy, you could use val1.clone() or Arrays.copyOf():

int[] val2 = Arrays.copyOf(val1, val1.length);

Objects (including instances of collection classes, String, Integer etc) work in a similar manner, in that assigning one variable to another simply copies the reference, making both variables refer to the same object. If the object in question is mutable, then subsequent modifications made to its contents via one of the variables will also be visible through the other.

Primitive types (int, double etc) behave differently: there are no references involved and assignment makes a copy of the value.

edited Sep 8, 2011 at 15:13

answered Sep 8, 2011 at 15:05

NPE

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Comments

Brian Knoblauch · Accepted Answer · 2011-09-08 15:03:00Z

4

Simply put, "val1" and "val2" are pointers to the actual array. You're assigning val2 to point to the same array as val1. Therefore, change one, and the other sees the same change. To have it truly be a copy, you'd have to clone the array instead of assigning.

answered Sep 8, 2011 at 15:03

Brian Knoblauch

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Comments

Rostislav Matl · Accepted Answer · 2011-09-08 15:49:47Z

4

Because arrays in Java are objects, i.e. passed by reference.

answered Sep 8, 2011 at 15:49

Rostislav Matl

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talnicolas · Accepted Answer · 2011-09-08 15:04:11Z

1

Because you assign a reference of val1 to val2, so they both point to the same object in the memory.

answered Sep 8, 2011 at 15:04

talnicolas

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