I always use constuctor only for attributes of one object.
But i thin when i wrote this :
public Predmet(int esp,int obaveze,int cena){
this.cena=cena;
this.obaveze=obaveze;
this.esp=esp;
List j = new ArrayList();
j.add(8);
this.nesto=(int) j.get(0);
}
where are stored this ArrayList,does it part of object,or is on stack and have reference to array in heap?
The ArrayList is created on the heap and only referenced by the local variable j from the stack here. After the execution of the constructor it will be eligible for garbage collection.
Constructor is working very similar the way methods work. At runtime, anything you define inside constructor/method are called local variables. Their scope will end as soon as execution hits the end of the constructor/method. After consturctor, your list will be eligible for GC. However, this.nesto will still get value 8 as its primitive type.
First, please don't use single variable names! This is bad practice and makes files harder to read. Using a more descriptive name helps to promote code readability.
The way that List j is created, it only exists in the scope of the constructor. After your constructor, List j is no longer accessible. If you wanted it to be accessible after the constructor, have a field for the object. For example:
public class Example {
private int specialNumber;
private List<Integer> numberList;
/**
* Constructs a new Example Object
*/
public Example(int exampleNum){
// specialNumber can be accessed from a Getter method (getSpecialNumber)
this.specialNumber = exampleNum;
this.numberList = new ArrayList<Integer>();
this.numberList.add(exampleNum);
// numberList is a field of this Example now
List<Integer> disappearingList = new ArrayList<Integer>();
disappearingList.add(exampleNum);
// After this method finishes, disappearingList will be gone
}
// disappearingList is no longer accessible
/**
* Gets this Example's specialNumber value
* #return int this.specialNumber
*/
public int getSpecialNumber(){
return this.specialNumber;
}
/**
* Gets this Example's numberList
* #return List<Integer> this.numberList
*/
public List<Integer> getNumberList(){
return this.numberList;
}
}
There is probably a way to hook into some of the Java cleaning methods and pull it out, but that will get a little messy. If you want to be able to create an Object inside another Object, and use it after the constructor, it must be saved as a field.
Related
I would like to know what is the best practice to return 'updated' ArrayList?
For example, if I am adding in a new element, it seems that whether if I did or did not specify the return type (see the addNewA() and addNewB()), the ArrayList in my main() method will still be 'updated' nonetheless.
Should I or should I not specify the return type?
Currently in my client program, most of the methods, I have specified it as void (no return type) and while the overall program still does works as intended, thought I would like to get this clarified and make the necessary changes if necessary.
public class MyClient
{
public static ArrayList<Person> addNewA(ArrayList<Person> myArray)
{
Person jack = new Person("jack", 24);
myArray.add(jack);
return myArray;
}
public static void addNewB(ArrayList<Person> myArray)
{
Person ben= new Person("ben", 19);
myArray.add(ben);
}
public static void main(String[] args)
{
ArrayList<Person> personArray= new ArrayList();
addNewA(personArray); // will return me an array size of 1
addNewB(personArray); // will return me an array size of 2
}
}
In a case like this, you should not return the list and should make your method void.
Different languages have different conventions, but one of Java's is that methods that operate by modifying their arguments (or the object they're called on) should not return values. Returning a value implies to someone using your code that a different object is being returned, since otherwise there is no use in returning an object the caller already has1. A method that is void, on the other hand, couldn't possibly be returning a copied-and-extended list, so it's very clear that it's intended to operate by modifying the list that you give it in the first place.
(By the way, you should also just use List<Person>, and you should pay attention to the warning you get about using new ArrayList() instead of new ArrayList<>().)
1 There is a specific exception to this, called the fluent builder pattern, but it's not easily confused with general code like this.
In java (and most high level strict type languages) Objects are passed by reference and primitives passed by value.
When using the new keyword you create an object.
While primitives (like int, char, double ect) are passed by value (meaning that a copy of the value of the variable will be sent to the invoked function), Object types are passed by reference, meaning that the actual original object is passed to the function.
To sum up - since you are using object here (ArrayList), you don't need a return type since the original object is changing.
The following class has an array of integer
public class customers {
public static int ID[];
public customers(int ID[]) {
ID = new int[10];
ID[0] = 00245;
ID[1] = 76644;
// more
}
//getters and setters
The subclass is defined as follow
public class songs extends customers {
//bunch of fields
The issue rises when within my array of objects. To create it, the following constructor was needed
public songs(int ID, // bunch of fields {
super(ID[0]);
this.ID = ID[];
// bunch of fields
Here, the super() method throws me back an error, that int[] in customers cannot be defined as a simple int.
Same goes when populating my array :
arraylist.add(new songs(ID[0], ...)); // didnt paste other variables
ID[0] is considered a simple int and not a int[].
While I understand the error itself, I don't know what causes it nor how to make java use my array of customers within the arrayList of Object defined in songs.
Thanks in advance !
If you want to send an array through subclass constructor you must first have a non-static (instance) array field in your super class, like this:
private int[] ids;
Be noticed that in java the fields are usually defined in camel case format.
Also you have a syntax error in this line:
super(ID[0])
You are referencing the int parameter defined in songs constructor as if an array, that is not correct.
Your call on super(ID[0]); is wrong: it calls the constructor of your members class, sending to it an int rather than an int[] as specified by your constructor. Moreover, I believe this.ID = ID[]; is wrong as well: "ID[]" in this context doesn't represent anything.
Also, as mentioned, static is probably not the good approach: it means that all Objects of type "Members" will share the same one-and-unique attribute "ID[]"!
More code would help. Especially about your songs, and the "arraylist" you're talking about.
I understand that immutable means that it is an object that will not change state after it is instantiated. But in this line of code I dont see Final when the array values is declared.
Is this class immutable? Can anyone explain how to find out. Thanks
public class A {
private double[] values;
public double[] getValues(){
return values;
}
}
As other have written this object is considered to be mutable in its state. What it is immutable to is that you can not exchange the array it holds. But you can change the array's content (getValues()[0] = 10;).
To convert this to a immutable object you must use List instead of an array. With List you can use Collections' method unmodifiableList to convert a given list into a version you can savely expose to the outside. If the caller of getValues() uses add or remove on a unmodifiable list it will result into a UnsupportedOpertionException keeping your object save from being modified.
If you need to stick to arrays you need to provide a copy (System.arraycopy) or a clone (clone()) of the array.
Usually a object is considered to be immutable if you can not change its properties (including inherited properties from superclasses. This usually includes the properties values as well but this is a blurred definition.
For example if you have a class that holds a File instance which points to document file and this File instance can not be changed the class is considered to be immutable (the inforamtion it provides never changes) but the document it points to can be mutated and changed every time. So its a blurred line actually (remember in your example you can not change the array but the content of the array).
Yes the code pasted is not having any final keyword associated and has no immutable behavior.
Well i would like to bring forth some key guidelines related to writing immutable classes in java :
1.) Ensure the class cannot be overridden - make the class final, or use static factories and keep constructors private
2.) Make fields private and final
force callers to construct an object completely in a single step, instead of using a no-argument constructor combined with subsequent calls to setXXX methods (that is, avoid the Java Beans convention)
3.) Do not provide any methods which can change the state of the object in any way - not just setXXX methods, but any method which can change state
4.) If the class has any mutable object fields, then they must be defensively copied when they pass between the class and its caller
A a = new A();
a.getValues()[0] = 1.2;
This would work as long as values is not empty. You will however not be able to reassign values to a new array. That is: a.getValues() = new double[5]; will not work.
The class is not immutable, as I can change values, just not reassign it.
Here is a simple verification. the values are initialized to 1,2.
Using the getter and a reference, one is able to change the values inside the first item in the array after the object is created
public class A {
private double[] values;
public double[] getValues() {
return values;
}
public static void main(String[] args) {
A test = new A();
test.values= new double[]{1, 2};
double[] valuesref = test.getValues();
valuesref[0] = 10;
for (int i = 0; i < test.values.length; i++) {
System.out.println(test.values[i]);
}
}
}
This can be avoided if getValues() returns a copy of the array.
Is there a way to move the entire contents of an ArrayList to another instance of ArrayList in O(1)?
I.e.: only the reference to the backing array is passed from one instance to the other (elements are not copied one by one).
For example:
ArrayList<String> a = new ArrayList<>(Arrays.asList("A", "B", "C"));
ArrayList<String> b = new ArrayList<>();
a.moveContentsTo(b);
// 'a' is now empty, while 'b' contains everything that 'a' did before and 'a != b'
// It is desired that the 'moveContentsTo' method is O(1)
Even better, is there an ArrayList#swapContents(ArrayList) method?
Further explanation and use-case:
Further explanation 1: the references of 'a' and 'b' must not be exchanged. I am not looking for tmp = a; a = b; b = tmp; type of solutions.
Further explanation 2: The operation must be ~O(1) in time.
The use-case: This is useful when an object wants to encapsulate a list constructed outside:
public class A {
private ArrayList<String> items = new ArrayList<>();
/**
* This method takes the sole ownership of the contents. Whoever
* passed the list from the outside will not be able to modify
* contents of 'this.items' from outside the class.
*/
public AnImmutableObject(ArrayList<String> items) {
if (items != null) {
items.moveContentsTo(this.items);
}
}
/**
* Other collections that do not provide the 'move' functionality
* must be copied. If we just stored the reference to 'items' we
* would break encapsulation as whoever called the constructor
* still have write capabilities to the collection.
*/
public A(Collection<String> items) {
if (items != null) {
this.items.addAll(items);
}
}
public List<String> getItems() {
return Collections.unmodifiableList(items);
}
}
Notice that we want to avoid making a copy (to increase speed and decrease memory usage). The crucial bit is that the callee must lose the ability to modify the (now encapsulated) ArrayList.
#Lirik answer is greate +1. However, if you are looking for a real ArrayList#swapContents(ArrayList), here is how you can do it:
public static void swapContents(ArrayList<String> listA, ArrayList<String> listB)
{
List<String> temp = new ArrayList<String>(listA);
listA.clear();
listA.addAll(listB);
listB.clear();
listB.addAll(temp);
}
AFAIK, it's very not Java-like to keep track of "ownership" of references (at least on the programmer's side) which is why I doubt that you'll find the std::move()-like functionality that you want. It just isn't very commonly needed in Java.
I guess C++ needs to keep track of object ownership explicitly because there is no garbage collection.
I think that your best bet is to create a defensive copy in your constructor and save space by relying on copy constructors of immutable objects:
public class AnImmutableObject {
private final List<String> items;
/**
* Creates a new immutable object with the given list of items.
* Always deep copy from an outside source, because we can't know if the
* calling code will modify that collection later.
*/
public AnImmutableObject(Collection<String> items) {
// It's a constructor. We know items needs to be set.
// We are creating a new array list, using its constructor which deep
// copies the collection, and immediately wrapping it to make it truly
// immutable. We are guaranteed that no one will hold a reference to the
// mutable view of the new ArrayList.
this.items = Collections.unmodifiableList(new ArrayList<String>(items));
}
/**
* Creates a new immutable object with the same items as another.
* Copying the reference here is completely safe because we
* enforce the immutability of the items array.
*/
public AnImmutableObject(AnImmutableObject source) {
items = source.items;
}
public List<String> getItems() {
return items;
}
}
At this point, you can "pass the arrays around" (really share them) in O(1) between your own immutable objects:
ImmutableObject a = new ImmutableObject(Arrays.asList("A", "B", "C")); // O(n)
ImmutableObject b = new ImmutableObject(a); // O(1)
Hopefully, something like this can suit your purposes.
Another route you could go is use Guava's ImmutableList. Since these are immutable, you can safely copy the reference to the ImmutableList in a constructor.
The main approach is about making it safe for you to copy references to the lists rather than taking ownership over them.
This should do it:
ArrayList<String> a = new ArrayList<>(Arrays.asList("A", "B", "C"));
ArrayList<String> b = a;
a = new ArrayList<>();
Conceptually speaking, a is now empty and b contains what a contained before. There was a single assignment and no copying of data, which is about the fastest you can do it. Does that satisfy your requirement, or do you actually want a to still reference the same array except that the given array should now be empty?
Update
I don't believe that in C++ the time complexity for the move operation is O(1) either. It's also prudent to point out that "because classes in Java use reference semantics, there are never any implicit copies of objects in those languages. The problem move semantics solve does not and has never existed in Java." (see the answer by FredOverflow: C++ Rvalue references and move semantics)
Is there a way to move the entire contents of an ArrayList to another ArrayList so that only the reference to the backing array is passed from one to the other (i.e., so that elements are not copied one by one).
Given the above statement, then if you copy something from array a to array b in Java, both arrays will reference the same data. All that you do with move semantics in C++ is that you save the temporary object which needs to be created in order to make such a copy:
X foo();
X x;
// perhaps use x in various ways
x = foo();
The last one does:
destructs the resource held by x,
clones the resource from the temporary returned by foo,
destructs the temporary and thereby releases its resource.
Move semantics does:
swap resource pointers (handles) between x and the temporary,
temporary's destructor destruct x's original resource.
You save one destruct, but only in C++... the above problem does not exist in Java! See this article for more details on move semantics: http://thbecker.net/articles/rvalue_references/section_02.html
Basically I have a variable, zlort = one;
I want to concatenate the value of zlort into a variable (object reference) name.
Like
BankAccount Accountzlort = new BankAccount;
I want the zlort in Account.zlort to actually be the replaced with value of zlort (one--meaning I want the value to be Accountone), and not zlort itself.
Is it possible to do this?
Thanks!
No you can't, but you might put the instance in a map:
Map<String,BankAccount> map = new HashMap<String,BankAccount>();
map.put("Account" + zlort, new BankAccount());
If you mean dynamically choosing the name to assign a variable to, then no.
You could use a HashMap to achieve the same effect.
It is not possible to change the name of a variable at runtime. That would lead to extreme security and stability problems when dealing with any real-world application.
However, as the two answers here have mentioned, a HashMap might acheive what you are looking for. (See the javadoc!!)
A HashMap (or any other map, for that matter) maps a Key to a Value. The concept is similar to a variable, which is a name -> value mapping. The only difference is that variables are part of the actual program code, which is effectively unmodifiable after compiling. A Map is a data structure that can be modified by the running program. This allows you to freely add key-value pairings to it.
Note that in Java, type-safety is encouraged through the use of Generics. Basically this ensures that the key can only be of one type (e.g. String) and the value can be of only one type (BankAccount). A thorough coverage of Generics can be found here.
You would declare this as follows:
Map<String, BankAccount> accounts = new HashMap<String, BankAccount>();
And then to add a key-value pair to the map, you would use the put() method (which 'puts' a value into the map, associated with a key)
String key = "Key"
BankAccount value = new BankAccount();
accounts.put(key, value);
To retrieve it, you would use the get() method.
BankAccount retrievedValue;
retrievedValue = accounts.get(key);
After reading the explanations in your comments, the fact that you can't use an array but can use an `ArrayList'...
Rather than creating a new variable name (or array element, or map value) for each BankAccount, you can probably use scope to your advantage.
Scope is the concept that a reference to a variable only has meaning within a certain part of code. If you declare a variable inside a method, that variable can only be seen within that method. A variable declared within a block (a loop, if statement, etc ) can only be seen from within that block.
Class fields have a different kind of scoping that can be adjusted with keywords (see here).
For example:
public class ScopeExample
int classInt = 10;
public void method() {
int methodInt = 0; // This integer can only be seen by code in
// this method
}
public void method2() {
//doSomething(methodInt) // This line won't compile because i is
// declared in a different method!
doSomething(classInt); // This line will compile and work
// because x is declared in the class that
// contains this method.
int index = 0;
while (index < 3) {
int whileInt = index; // This integer can only be seen from within
// this while loop! It is created each
// loop iteration.
doSomething(whileInt);
}
doSomething(whileInt); //This line won't work, whileInt is out of scope!
}
public doSomething(int a) {
System.out.println(a);
}
}
SO! If you create a BankAccount object within the loop, you don't have to worry about creating a new name for the next one. Each time the loop iterates it will become a new object (when you create it).
If you have to store it, you definitely will need to use an array or other data structure (ArrayList!).
Building on the idea of scope, you -can- have the same variable name for each new BankAccount. A variable reference name isn't guaranteed to be paired with the object that it refers to. That is a convenience to the programmer, so you don't have to know the exact memory address it is being stored in.
For example:
public static void main(String[] args) {
Object o;
int i = 0;
while (i < 5) {
Object reference = new Object(); // Create a new Object and store
// it in 'reference'
o = obj; // The Object 'o' now refers to the object in 'reference'
i++;
}
System.out.println(o); // This should print information about the
// LAST object created.
}
The new Object created in the loop does not belong to 'obj'. You as a programmer use 'obj' to point to the Object. The program doesn't really know what obj means, other than the fact that it points to the Object you just created.
Finally, you can use this along with an ArrayList to make your life easier.
public static void main(String[] args) {
// Our new ArrayList to hold our objects!
ArrayList<Object> stuff = new ArrayList<Object>();
int i = 0;
while (i < 5) {
Object obj = new Object(); // Create an object and make obj point to it.
stuff.add(obj); // Put "the object that 'obj' points to" in 'stuff'.
i++;
}
// This loop goes through all of the Objects in the ArrayList and prints them
for (int index = 0; index < stuff.size(); index++) {
System.out.println(stuff.get(i)); // This will print a single
// object in the ArrayList each time.
}
}