Cfxtrjtrk

I am a beginner in Java. When developing a program, I created an object with a constructor with variables as arguments. But when I change the value of the variable after creating the object, my object has the second value instead of the first one. I don't want my object to change the value. What do I do?

public class Person {



    public Person(int arrayTest) {

            this.arrayTest = arrayTest;

    }



    public int getArray() {

        return this.arrayTest;

    }



    public boolean canHaveAsArray(int arrayTest) {

            return true;

    }



    private int arrayTest = new int[2];



    public static void main(String args) {

        int array = new int {5, 10};

        Person obj1 = new Person(array);

        array[0] = 20;

        System.out.println(Arrays.toString(obj1.getArray()));

    }

}

My output should be [5, 10], but instead, I am getting [20,10]. I need to get [5,10] even when I change an element of the array as shown above. What should I do?

Array is passed by reference in Java. If you pass the original array to the constructor of Person, you are passing the reference to the original array. So any change in arrayTest inside Person instance will reflect in original array(int array) and vice-versa.

If you don't want to change the value of elements of original array in Person instance then you have two options:

• 
  

  You can modify the code in Person constructor to create a copy of original array using java.util.Arrays.copyOf method and then use that copy:

  
  
  

  public Person(int arrayTest) {
  
      this.arrayTest = java.util.Arrays.copyOf(arrayTest, arrayTest.length);
  
  }
  
  

  
  


• 
  

  Don't pass the original array to constructor, instead just send a copy of original array:

  
  
  

  Person obj1 = new Person(java.util.Arrays.copyOf(array, array.length));
  
  

  
  

However, I would prefer first approach.

There is no such thing as immutable (unchangeable) array in Java. The Java language does not support this, and neither does the JVM. You can't solve this at the language level.

In general, the only way to prevent changes to an array is to not share the reference to the array with other code that might change it.

In your example, you have what is known as a leaky abstraction. You are passing an array to your Person class, and the caller is keeping a reference to that array so that it can change it. To solve this, you can:

• copy the array, and pass a reference to the copy, or


• have the constructor (or a setter for the array attribute) make the copy.

(See answer https://stackoverflow.com/a/55428214/139985 for example code.)

The second alternative is preferable from an OO perspective. The Person class should be responsible for preserving its own internal state from interference ... if that is your design requirement. It should not rely on the caller to do this. (Even if the caller is technically part of the same class as is the case here.)

There is no unmodifiable array, but you can make an unmodifiable list:

List<Integer> list = List.of(5, 10);

You will have to change your code to use lists instead of arrays, but this is generally preferable anyway.

If you already have an array of a non-primitive type, you can wrap it in an unmodifiable list, like so:

List<Integer> list = Collections.unmodifiableList(Arrays.asList(array));

However, while you can't change the list directly, changing the array will change the list. Moreover, this won't work on int, but only on subclasses of Object.

Instead of passing a copy of the array to the object, as others have suggested, I would recommend that the Person object's constructor should create a copy. Which means instead of,

this.arrayTest = arrayTest;

It should be

this.arrayTest = Arrays.copyOf(arrayTest, arrayTest.length);

This would allow the object to be defensive against malicious code trying to modify arrays after construction and validation by constructor. In fact most IDEs have analysis tools which will give you a warning against saving array reference.

In Java, objects/arrays are manipulated through reference variables^#

When a function is invoked with arrays as their arguments, only a reference to the array is passed. Therefore, when you mutate array array, the arrayTest field also get mutated as they are referring to the same address

To override this behavior, you can create a copy of the array in your constructor using Object.clone() method like:

public Person(int arrayTest) {

    this.arrayTest = arrayTest.clone();

}

_{# Source: Wikipedia}

As others have already pointed out: The array is passed as a reference to the Person. So changes that are later done to the array will be visible to the Person object. But that's only one half of the problem: You are not only passing a reference to the array to the constructor of the Person, you are also returning a reference from the getArray method.

Generally speaking, and as StephenC already pointed out in his answer: One important aspect of Object-Oriented design is to properly manage the state space of objects. It should not be possible for users of a class to bring an object into any form of "inconsistent state".

And this is difficult with plain primitive arrays. Consider the following pseudocode, referring to the class that you posted:

int originalArray = new int[2];

originalArray[0] = 12;

originalArray[1] = 34;



Person person = new Person(originalArray);

int arrayFromPerson = person.getArray();



originalArray[0] = -666;               // Modify the original array

System.out.println(arrayFromPerson[0]) // Prints -666 - this is unexpected!



arrayFromPerson[1] = 12345678;         // Modify the array from the person

System.out.println(originalArray[1])   // Prints 12345678 - this is unexpected!

Nobody knows who has a reference to the array, and nobody can verify or track that the contents of the array is not changed in any way. How critical this is becomes more obvious when you anticipate that the Person object will be used at different places, possibly even by multiple threads.

Plain primitive arrays in Java do have their justification. But when they appear in the interface of a class (that is, in its public methods), they should be view with scrutiny.

In order to be absolutely sure that nobody can interfere with the array that is stored in the Person object, you'd have to create defensive copies everywhere:

public Person(int arrayTest) {

    this.arrayTest = arrayTest.clone(); // Store a clone of the array

}

public int getArray() {

    return this.arrayTest.clone(); // Return a clone of the array

}

But this may be cumbersome. A more object-oriented solution could be to expose a "read-only view" on the state that is represented with the array. For example:

public Person(int arrayTest) {

    this.arrayTest = arrayTest.clone(); // Store a clone of the array

}

public int getArrayLength() {

    return this.arrayTest.length;

}

public int getArrayElement(int index) {

    return this.arrayTest[index];

}

^{(Of course, in practice, you'd name the methods accordingly, depending on what the array actually represents. For example, if it's the ages of the children of the person, you'd call the methods getNumChildren() and getAgeOfChild(int i) or so...)}

Another option how this can be solved is to expose an (unmodifiable) List view on the array. This can, for example, be done with the asUnmodifiableList method that is shown in this answer.