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How to prevent changing the value of variable?



The Next CEO of Stack OverflowIs Java “pass-by-reference” or “pass-by-value”?How do I efficiently iterate over each entry in a Java Map?Sort a Map<Key, Value> by valuesHow do I call one constructor from another in Java?How do I read / convert an InputStream into a String in Java?How do I generate random integers within a specific range in Java?How to get an enum value from a string value in Java?How do I determine whether an array contains a particular value in Java?How do I convert a String to an int in Java?How do I fix android.os.NetworkOnMainThreadException?










7















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?










share|improve this question


























    7















    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?










    share|improve this question
























      7












      7








      7








      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?










      share|improve this question














      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?







      java






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 42 mins ago









      OpheliaOphelia

      362




      362






















          2 Answers
          2






          active

          oldest

          votes


















          5














          Array is passed by reference in Java. If you pass the original array to the constructor of Person, you are just passing the reference to the original array and the changes in original array will reflect in Person instance.



          So if you don't want to change the value of array in Person so don't pass the original array, instead just send a copy of original array like below:



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


          You can also modify the code in Person constructor to achieve the same results:



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






          share|improve this answer
































            3














            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 case, you have what is known as a "leaky abstraction". You are passing an array to your Person class, and keeping a reference to that array so that you 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.)






            share|improve this answer

























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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5














              Array is passed by reference in Java. If you pass the original array to the constructor of Person, you are just passing the reference to the original array and the changes in original array will reflect in Person instance.



              So if you don't want to change the value of array in Person so don't pass the original array, instead just send a copy of original array like below:



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


              You can also modify the code in Person constructor to achieve the same results:



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






              share|improve this answer





























                5














                Array is passed by reference in Java. If you pass the original array to the constructor of Person, you are just passing the reference to the original array and the changes in original array will reflect in Person instance.



                So if you don't want to change the value of array in Person so don't pass the original array, instead just send a copy of original array like below:



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


                You can also modify the code in Person constructor to achieve the same results:



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






                share|improve this answer



























                  5












                  5








                  5







                  Array is passed by reference in Java. If you pass the original array to the constructor of Person, you are just passing the reference to the original array and the changes in original array will reflect in Person instance.



                  So if you don't want to change the value of array in Person so don't pass the original array, instead just send a copy of original array like below:



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


                  You can also modify the code in Person constructor to achieve the same results:



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






                  share|improve this answer















                  Array is passed by reference in Java. If you pass the original array to the constructor of Person, you are just passing the reference to the original array and the changes in original array will reflect in Person instance.



                  So if you don't want to change the value of array in Person so don't pass the original array, instead just send a copy of original array like below:



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


                  You can also modify the code in Person constructor to achieve the same results:



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







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 29 mins ago

























                  answered 34 mins ago









                  Aniket SahrawatAniket Sahrawat

                  6,32121339




                  6,32121339























                      3














                      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 case, you have what is known as a "leaky abstraction". You are passing an array to your Person class, and keeping a reference to that array so that you 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.)






                      share|improve this answer





























                        3














                        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 case, you have what is known as a "leaky abstraction". You are passing an array to your Person class, and keeping a reference to that array so that you 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.)






                        share|improve this answer



























                          3












                          3








                          3







                          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 case, you have what is known as a "leaky abstraction". You are passing an array to your Person class, and keeping a reference to that array so that you 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.)






                          share|improve this answer















                          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 case, you have what is known as a "leaky abstraction". You are passing an array to your Person class, and keeping a reference to that array so that you 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.)







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited just now

























                          answered 18 mins ago









                          Stephen CStephen C

                          525k72585944




                          525k72585944



























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