Clone method for Java arrays

When the clone method is invoked upon an array, it returns a reference to a new array which contains (or references) the same elements as the source array.

So in your example, int[] a is a separate object instance created on the heap and int[] b is a separate object instance created on the heap. (Remember all arrays are objects).

    int[] a = {1,2,3};
    int[] b = a.clone();

    System.out.println(a == b ? "Same Instance":"Different Instance");
    //Outputs different instance

If were to modify int[] b the changes would not be reflected on int[] a since the two are separate object instances.

    b[0] = 5;
    System.out.println(a[0]);
    System.out.println(b[0]);
    //Outputs: 1
    //         5

This becomes slightly more complicated when the source array contains objects. The clone method will return a reference to a new array, which references the same objects as the source array.

So if we have the class Dog...

    class Dog{

        private String name;

        public Dog(String name) {
            super();
            this.name = name;
        }

        public String getName() {
            return name;
        }

        public void setName(String name) {
            this.name = name;
        }

    }

and I create and populate an array of type Dog...

    Dog[] myDogs = new Dog[4];

    myDogs[0] = new Dog("Wolf");
    myDogs[1] = new Dog("Pepper");
    myDogs[2] = new Dog("Bullet");
    myDogs[3] = new Dog("Sadie");

then clone dog...

    Dog[] myDogsClone = myDogs.clone();

the arrays refer to the same elements...

    System.out.println(myDogs[0] == myDogsClone[0] ? "Same":"Different");
    System.out.println(myDogs[1] == myDogsClone[1] ? "Same":"Different");
    System.out.println(myDogs[2] == myDogsClone[2] ? "Same":"Different");
    System.out.println(myDogs[3] == myDogsClone[3] ? "Same":"Different");
    //Outputs Same (4 Times)

This means if we modify an object accessed through the cloned array, the changes will be reflected when we access the same object in the source array, since they point to the same reference.

    myDogsClone[0].setName("Ruff"); 
    System.out.println(myDogs[0].getName());
    //Outputs Ruff

However, changes to the array itself will only affect that array.

    myDogsClone[1] = new Dog("Spot");
    System.out.println(myDogsClone[1].getName());
    System.out.println(myDogs[1].getName());
    //Outputs Spot
    //        Pepper

If you generally understand how object references work, it is easy to understand how arrays of objects are impacted by cloning and modifications. To gain further insight into references and primitives I would suggest reading this excellent article.

Gist of Source Code