How does x =new int[2]; work in this java code?

Question

Analyze the following code:

class Test {

    public static void main(String[] args) {
        int[] x = {1, 2, 3, 4};
        int[] y = x;

        x = new int[2];
        for (int i = 0; i < y.length; i++) {
            System.out.print(y[i] + " ");
        }
    }
}

• A. The program displays 1 2 3 4
• B. The program displays 0 0
• C. The program displays 0 0 3 4
• D. The program displays 0 0 0 0

The answer for the following code is A. But why the answer is not B ?

Answer 1

Let's assume that {1, 2, 3, 4} has memory address M.

When assigning x to {1, 2, 3, 4}, you're assigning a reference to the memory address of {1, 2, 3, 4}, i.e. x will point to M.

When assigning y = x, then y will refer M.

After that, you're changing the reference where x points to, let's say it's N.

So, when printing, y points to M (which is the address of {1, 2, 3, 4}), but x holds reference to N (which is new int[2]). And here comes the difference.

Answer 2

To give you a more clear explanation:

int[] x = {1, 2, 3, 4};       // step 1

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int[] y = x;    // step 2

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 x = new int[2];        // step 3

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In the third step, when the x changes, y is not affected because you are changing the reference of x, not the address of the array. So it doesn't affect the value of y.

Answer 3

Because int[] y = x; the array y will have some variables like this: {1,2,3,4}.

This line x = new int[2]; will creating a different space for x pointing to it.

So the result will be {1,2,3,4} of the array y