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?
Is 'diverse range' a pleonastic phrase?
Are there any limitations on attacking while grappling?
How to count occurrences of text in a file?
What's the best way to handle refactoring a big file?
Received an invoice from my ex-employer billing me for training; how to handle?
Is micro rebar a better way to reinforce concrete than rebar?
What can we do to stop prior company from asking us questions?
How to prevent changing the value of variable?
Real integral using residue theorem - why doesn't this work?
What is "(CFMCC)" on an ILS approach chart?
What does "Its cash flow is deeply negative" mean?
Why am I allowed to create multiple unique pointers from a single object?
Is there a difference between "Fahrstuhl" and "Aufzug"
"In the right combination" vs "with the right combination"?
Contours of a clandestine nature
Inappropriate reference requests from Journal reviewers
Why has the US not been more assertive in confronting Russia in recent years?
Is it possible to search for a directory/file combination?
Multiple labels for a single equation
Make solar eclipses exceedingly rare, but still have new moons
What benefits would be gained by using human laborers instead of drones in deep sea mining?
Interfacing a button to MCU (and PC) with 50m long cable
Why does the UK parliament need a vote on the political declaration?
If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?
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?
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
add a comment |
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
add a comment |
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
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
java
asked 1 hour ago
OpheliaOphelia
362
362
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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);
add a comment |
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.)
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55428172%2fhow-to-prevent-changing-the-value-of-variable%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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);
add a comment |
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);
add a comment |
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);
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);
edited 49 mins ago
answered 54 mins ago
Aniket SahrawatAniket Sahrawat
6,32121339
6,32121339
add a comment |
add a comment |
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.)
add a comment |
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.)
add a comment |
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 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.)
edited 19 mins ago
answered 38 mins ago
Stephen CStephen C
525k72585944
525k72585944
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55428172%2fhow-to-prevent-changing-the-value-of-variable%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown