5 
public class Test {
    public static void main(String[] args) {
        int[] a = {1, 2, 3, 4, 5};
        int[] b = a;
        int[] c = {6, 7, 8};
        a = c;
        for(int i = 0; i < a.length; i ++)
            System.out.print(a[i] + " ");
        System.out.print("\n");
        for(int i = 0; i < b.length; i ++) 
            System.out.print(b[i] + " ");
        System.out.print("\n");
    }
}

I have initialized array a and assigning reference of a to new array b. Now I initialized a new array c and passed its reference to array a. Since array b is reference to array a, b should have new values that are in c but b is having old values of a. What is the reason behind it? Output is given below -

Output -

6 7 8 
1 2 3 4 5
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  • Thats fine this is how each programming language works. Commented Jun 24, 2017 at 4:27

5 Answers 5

Reset to default
10

Don't be irritated by the name 'list'. The images are taken from a python visualization, but the principle is the same in Java

Array a is assigned with a new array:

Assigning a

Array b is assigned to the instance behind a:

Assigning b

Array c is assigned with a new array:

Assigning c

And finally a is assigned to the instance behind c, b was not re-assigned and still keeps a reference to the old a:

Re-Assigning a

Images taken from a visualization on pythontutor.com

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Comments

1

Suppose you think of an object as a house, and a reference as a piece of paper with the address of a house written on it. a, b, and c are all references. So when you say

    int[] a = {1, 2, 3, 4, 5};

you're building a house with 1, 2, 3, 4, and 5 in it; and a is a piece of paper with the address of that house. 

    int[] b = a;

b is another reference, which means it's another piece of paper. But it has the address of the same house on it.

    int[] c = {6, 7, 8};
    a = c;

Now we build a new house and put 6, 7, and 8 into it. c will be a piece of paper with the address of the new house. When we say a = c, then the slip of paper that used to be a is thrown out, and we make a new piece of paper with the address of the new house. That's the new value of a.

But b was a separate piece of paper. It hasn't changed. Nothing we've done has changed it.

References are like that.

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1

When you assigned value of a to b, it means b is referring to same space allocated to array a. This means b will pick up any changes made to the array a, but if any changes made to the variable a. If a is made to refer to new array, b will still refer the old a reference.

b = a;  // b basically referring to memory used by array a
a = c;  // reference pointed by a has changed, but b is still referring the old array a
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When you do b=a b references to a. However when you to a=c a is referring to c, but still b refers to the old object (address of this object as value been assigned and it is constant) a that you assigned because that is what it contains when you assigned.

Unless you reassign it, it won't change.

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This is why:

What is Object Reference Variable?

You have 2 objects (the arrays) and 3 references (a, b, c) to the objects. You're just swapping around where things are pointed.

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