How to prevent changing the value of variable?
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 53 mins 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 |
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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 40 mins ago
answered 45 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 10 mins ago
answered 29 mins ago
Stephen CStephen C
525k72585944
525k72585944
add a comment |
add a comment |
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