I am trying to implement a copy constructor in java. I am facing a problem with non primitive type fields of the class. While creating a new copy, it is sharing the members. For Example
public class Bad implements Cloneable {
private ArrayList<Integer> a;
private Object c;
public static void main(String[] args) {
Bad b1 = new Bad();
b1.a.add(10);
System.out.println(b1.a);
Bad b2 = b1.clone();
b2.a.add(12);
System.out.println(b1.a);
}
Bad() {
a = new ArrayList<>();
c = null;
}
Bad(Bad b) {
a = b.a;
c = b.c;
}
public Bad clone() {
return new Bad(this);
}
}
And the result is :
[10]
[10, 12]
I don't want this to happen. Take this as an example. My original problem consists of even more fields that are user defined.
Or are there any libraries which do the work for me? Thanks in advance.
Simple rules for a copy constructor:
primitive values may be copied as-is; they are just values without a separate identity
references to immutable types (eg. String, Integer, any enum class constant) may also be copied as-is; although the original and the copied objects will share the same reference, the referred object is immutable and will never change
references to mutable types (eg. Date, ArrayList, any array) must be copied to a new instance of the type; otherwise the original and the copied object will share a reference to the same mutable field object (which is not what you want)
Making a copy of an object that contains only fields with primitive and immutable values is easy mode.
Copying an object whose fields contain mutable objects can make the process arduous and expensive depending on how complex the mutable object is (imagine an ArrayList that contains a Map whose values are also Maps). Making a new copy of the mutable field is, however, essential if you wish to have a safe copy.
Integer is immutable but you need to create a totally new ArrayList, and I mean here :
Bad(Bad b) {
a = b.a;
c = b.c;
}
do instead
Bad(Bad b) {
a = new ArrayList<>(b.a);
c = // this here must be copy constructed too
}
and then you will get
[10]
[10]
The correct way would be creating a new instance of the list instead of passing a reference to the original list.
Bad(Bad b) {
a = new ArrayList<>(b.a);
c = b.c; // this should call clone or something similar as well
}
Also note that if you would have some non-primitive type within the list of b.a, then you would have to copy/clone all the sub-elements as well (it is not needed now as you have Integer in it which is not mutable).
Related
Basically I got told that I shouldn't have public methods returning Arrays for "future" security purposes. Instead they should be private, and if I wanna return an Array it would have to be some kind of copy in another method.
This is how it looks now..
public Object[] ownedObject() {
return objectArr;
}
If I make this private the class that needs it doesn't recognize the method above.
Thing is I need to use the contents in that Array in said, other class, and the total project, as I have it right now with 5 different classes, works (with the returning Array-methods set to Public and not private).
As you are concerned with security aspect of this problem you might want to distinguish between shallow copy and deep copy of the array. If your array contains mutable objects you probably need a deep copy of every single element in the array to ensure that state is not leaking from your object.
Assuming that you array is of type MyType with a copy constructor:
public MyType[] ownedObject() {
MyType[] copyArr = new MyType[objectArr.lenght];
for (int i = 0; i < objectArr.lenght; i++) {
copyArr[i] = new MyType(objectArr[i]);
}
return copyArr;
}
There are also other ways to deep copy an object.
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?
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.
If you would like to prevent the value of variable which is of primitive type, you can do so using final keyword. Eg:
private final int test = 1;
To prevent changing the value inside an object you can mark the fields as final. A final keyword in declaration of object instance means the variable can't be reassigned and doesn't guarantee that the object state won't change if the reference to that object is shared. To prevent changing the state of a particular object, you should mark it's field as final.
There is no such thing as immutable (unchangeable) array in Java. The Java language does not support this. As JLS 4.12.4 states:
If a final variable holds a reference to an object, then the state of the object may be changed by operations on the object, but the variable will always refer to the same object. This applies also to arrays, because arrays are objects; if a final variable holds a reference to an array, then the components of the array may be changed by operations on the array, but the variable will always refer to the same array.
The JVM spec doesn't support an immutable array type either. You can't solve this at the language level. The only way to avoid 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[].
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
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.
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.
as you are beginner in java you write following code in constructor but it is better to use clone method as marco13 and rv 7 already explained
and as sourabh bhat explained we can also use Arrays class copyof
the idea behind all of above logic is simple don't pass referance of current object but create clone of object and pass that clone or just copy each content of the object
public Person(int[] arrayTest) {
for (int i = 0; i <this.arrayTest.length; i++) {
this.arrayTest[i]=arrayTest[i];
}
}
Here is what happens in memory:
Program: Stack memory: Heap memory:
int[] array = new int[] {5, 10}; array -> 0x77a89 0x77a89 {5, 10}
Person obj1 = new Person(array); obj1.arrayTest -> 0x77a89 No change
array[0] = 20; 0x77a89 {20, 10}
As you can see stack memory holds only the address of the object which gets created in the heap memory. So when you change the arrays value it automatically changes in the Person obj1 object as well.
To fix this you need to create a new Object in memory so that the actual Objects value is copied. To do this we can:
[1] Use the clone property of array.
public Person(int[] arrayTest) {
this.arrayTest = arrayTest.clone();
}
[2] Or we can create our own clone.
public Person(int[] arrayTest){
if (arrayTest == null){
this.arrayTest = null;
} else {
int[] copyArray = new int[arrayTest.length];
for(int i=0; i<arrayTest.length; i++) {
copyArray[i] = arrayTest[i]
}
this.arrayTest = copyArray;
}
}
Either way a new Object is created in memory and this prevents the object from being shared.
Edit 5/5/19: Source code added
As most answers point out, there is no immutable array of primitives in Java. So you have to do some tricks.
Pure Java: Make defensive copies. Most answers show how to store a copy of the array that is received as a parameter in the constructor. But only one answer mentions that you also have to return a copy of the internal array with getArray().
public class Person {
final private int[] arrayTest;
public Person(int[] arrayTest) {
this.arrayTest = java.util.Arrays.copyOf(arrayTest, arrayTest.length);
}
public int[] getArray() {
return java.util.Arrays.copyOf(arrayTest, arrayTest.length);;
}
}
Other internal representation: Store the array as a (mutable) ArrayList, which is based on an array and should have best performance. You have to convert from array to List in the constructor and from List to array in getArray(). There is no need to use Collections.unmodifiableList() (or Guavas ImmutableList<>) as long as you write no method that could modify the List because no one will have access to the List.
public class Person {
final private List<Integer> arrayTest;
public Person(int[] arrayTest) {
this.arrayTest = new ArrayList<>(Arrays.asList(arrayTest));
}
public int[] getArray() {
return this.arrayTest.stream().mapToInt(Integer::valueOf).toArray;
}
}
Let other people do the job. Google AutoValue auto-generates immutable classes. And provides equals(), hashCode() and toString(). Easy to use. My favorite solution.
import com.google.auto.value.AutoValue;
#AutoValue
public abstract class Person {
public static create(int[] arrayTest) {
return new AutoValue_Person(int[] arrayTest);
}
public abstract int[] getArray() {}
}
Suppose I have an array
public static final String[] fooArray ={ Foo.a, Foo.b, Foo.c };
where Foo.a b and c are static final Strings.
Could I still do something like fooArray[0] = "taco"; and end up with { taco, Foo.b, Foo.c } as the contents of fooArray?
If so, would making the array private, and having a getter that makes a copy of the array using Arrays.copyOf solve this issue?
The final applies to the array reference, not its entries. Different strings can still be written to its entries.
If so, would making the array private, and having a getter that makes a copy of the array using Arrays.copyOf solve this issue?
Yes, defensive copies are a fairly standard way to handle this.
Alternately, given what you've outlined, you don't need to have the array at all, just a getter that looks like this:
public String[] getFooArray() {
return new String[] { Foo.a, Foo.b, Foo.c };
}
Or as jtahlborn commented, use an unmodifiable List<String>:
public static final List<String> fooArray;
static {
List<String> a = new ArrayList<>();
Collections.addAll(a, Foo.a, Foo.b, Foo.c);
fooArray = Collections.unmodifiableList(a);
}
// (There's probably some really nifty Java8 way to do that as a one-liner...
Yes.
A final array means you can't reassign the array.
So you couldn't do: fooArray = new String[]{...};.
But you can however change what is inside the array. This is the effect of saying: "You can't change the box, but you can change the apples inside the box to be oranges." The box stays the same, and is final, but you've effectively changed the contents.
That being said, if you encapsulate the class, then you can just clone the array when it is needed.
This is currently employed by many classes, such as String#toCharArray and Enum#values, where changing the array's contents comprises the integrity of the finalized object(s).
The final-modifier will only prevent changing the fooArray-reference, not the contents of the array itself. Making it private and having a getter returning a copy would hide the original array, and any changes made to the returned array would only affect the copy. However, it would still be possible to modify the original via reflection, but if your intent is to only prevent accidental modification of the original array, that would work.
Rest have answered about the final well. Just a suggestion on the other part - rather than implementing a getter which does a copy of entire array, if your scenario allows, its better to have a getArrayElement(int position) where you just return an array element rather than the whole array - copying an array is expensive.
You could make a getter that returns a mutable copy of an immutable value.
If you used array copy the values inside the copy will still be final.
public class HelloWorld{
public static void main(String []args){
System.out.println("Hello World");
final int b = 5;
int c = b; // Because of c being mutable we can now change this copy
c = 7;
System.out.println(c);
}
}
... Some psudo code -> for copying an iterable into a mutable form.
public collection<int>(final collection<final int> finalCollection )
collection nonFinalCollection = new collention();
for(k : finalCollention){collection.add((int) k)}
return(collection)
This is my Class:
public class City
{
private String _cityName;
private Point _cityCenter;
private Point _centralStation;
private long _numOfResidents;
private int _noOfNeighborhoods;
private final long RESIDENTS_CONST_VALUE = 0;
private final int NEIGHBORHOODS_CONST_VALUE = 1;
}
One of my constructor is Copy from other object (same object):
public City(City other)
{
_cityName = other._cityName;
_cityCenter = other._cityCenter;
_centralStation = other._centralStation;
_numOfResidents = other._numOfResidents;
_noOfNeighborhoods = other._noOfNeighborhoods;
}
public Point(Point point)
{
_x = point._x;
_y = point._y;
}
This constructor get another City object and copy it's values.
My question is if what i have done it's OK to avoid aliasing or i need to do something else
The problem is that you are copying the reference from old object to new.It will create problems while copying mutable fields.If both of the objects share common reference to a field, changing the value in one object will affect copied object as well.
The only problem I see is with the reference to the Point class (I'm assuming that we are taking about java.awt.Point). This class is mutable, so the City class you are copying from can change it, and the change will be reflected in your copy also. Use the following code to copy the Point object:
_cityCenter = new Point(other._cityCenter);
_centralStation= new Point(other._centralStation);
The rest of the fields are either primitives or immutable, so it is OK
What you have done looks sane, at least because you haven't copied the Point objects, those are most likely mutable, i.e. if you copy them by simple assignment you would copy only the reference and changes to one object will reflect in the 'copy' -> shallow copy. If you need to make a copy of them too then you'll have to implement a "copy constructor" for them too.
However the standard way of doing this in java is to implment the Colneable interface and override the clone method.
I am a little confused as to how my Java program is allocating memory. Below is a typical situation in my program. I have an object created on the Heap inside of a class which I pass to other functions inside of other classes.
public class A {
private List<Integer> list = new ArrayList<Integer>();
public A() {
for(int i = 0; i < 100; i++)
list.add(i);
}
public void foo() {
B b = new B();
b.bar(this.list);
}
}
public class B {
A a = new A();
public int bar(List<Integer> list) {
list.add(-1);
return list.size();
}
public int getSize() {
List<Integer> list = a.list;
return a.size();
}
}
My question(s) are about how the memory is used. When A's list gets passed to B.bar(), does the memory get copied? Java is pass-by-value so I assumed the B.bar() function now contains a copy of the list which is also allocated on the heap? Or does it duplicate the list and put it on the stack? Or is the list duplicated at all? Similarly what applies to these memory questions inside of B.getSize()?
It's probably clearer to say that in Java, object references are passed by value.
So B gets a copy of the reference (which is a value) to the List, but not a copy of the List itself.
You can easily prove this to yourself by getting B to modify the List (by adding or removing something), and observing that A's List also changes (i.e. it is the same List).
However, if you change B's reference to the List (by creating and assigning a new List to it), A's reference is unchanged.
Java is pass-by-value, but the value of an expression or variable is always either a primitive or a reference. A reference, in turn, is either null or a pointer to some object.
In your case: private List<Integer> list = new ArrayList<Integer>(); declares list as having a reference type, and makes its value a pointer to a newly allocated ArrayList object.
When you use list as an actual argument, for example on the b.bar(this.list); call, the formal argument list in b.bar is initialized with a copy of that pointer. That is, bar's formal argument points to the same ArrayList object as A created. The ArrayList object is not copied.
Java is pass-by-value, dammit!
However, you only ever have access to references to Objects. In that sense, you are passing Object references by value.