what is happening when try to use array variable

Question

public class Demo {
    public static void main(String[] args) {
        int array[]=new int[2];
        System.out.println(array);
    }
}

This an example. Can anybody explain this.

What don't you understand exactly? – Alexis C. Nov 29 '14 at 16:36 — Alexis C., Nov 29 '14 at 16:36
What did happen when you ran the code? – reto Nov 29 '14 at 16:37 — reto, Nov 29 '14 at 16:37

score 0 · Answer 1 · answered Nov 29 '14 at 16:39

You just defined the array and java by default initialized that array context i.e. array[0] and array[1] with default value i.e. 0.

When you will print the array it will just print the type of Array i.e. Int array followed by hashCode something like I@32324.

score 0 · Accepted Answer · edited May 23 '17 at 10:26

// declares an array of integers
int[] anArray;

// allocates memory for 10 integers
anArray = new int[10];

//declares an array of integers and allocates memory for 2 integers
int array[]=new int[2];

And if you want to print that thing use --> this <--

//print the content of the array
System.out.println(Arrays.toString(array));

Because this :

System.out.println(array);

is not going to print the content of your array, it will print the memory address of the array.

what is happening when try to use array variable

2 Answers2