What exactly does the clone() method in Java return when used on an array? Does it return a new array with data copied from the original?
Ex:
int[] a = {1,2,3}; int[] b = a.clone();
Java allows you to copy arrays using either direct copy method provided by java. util or System class. It also provides a clone method that is used to clone an entire array.
The Java Object clone() method creates a shallow copy of the object. Here, the shallow copy means it creates a new object and copies all the fields and methods associated with the object. The syntax of the clone() method is: object.clone()
clone() method in Java Object class has a clone method which can be used to copy the values of an object without any side-effect. Assignment operator has a side-effect that when a reference is assigned to another reference then a new object is not created and both the reference point to the same object.
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
clone()
method creates and returns a copy of this object. The precise meaning of "copy" may depend on the class of the object. The general intent is that, for any object x, the expression:
x.clone() != x
Will be true, and that the expression:
x.clone().getClass() == x.getClass()
Will be true, but these are not absolute requirements.
While it is typically the case that:
x.clone().equals(x)
will be true, this is not an absolute requirement.
By convention, the returned object should be obtained by calling super.clone
. If a class and all of its superclasses (except Object) obey this convention, it will be the case that x.clone().getClass() == x.getClass()
.
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