I am trying to understand generics in Java.
private List<Own> l = new ArrayList<Own>();
I have the following error :
no instance of Typed array variable T exist so that List<Own> conform to T[]
when I pass it in a method (readTypedArray) that expects T[].
private List<Own> list = new ArrayList<Own>();
private OwnParceable(Parcel in) {
in.readTypedArray(list, CategoriesParceable.CREATOR);
}
The method in.readTypedArray() expects an array T[], but you passed a List<Own which is not an array.
List is not an array you can't use it where an array is expected, List is an interface which extends Collection while array is a data structure in Java, check Difference between List and Array for further details.
You can either declare an Own[]instead of List<Own> or convert this list into an array before passing it to the method, check Convert list to array in Java:
in.readTypedArray(list.toArray(new Own[list.size()]), CategoriesParceable.CREATOR);
This has nothing to do with generics - Lists and arrays are just two different things. If your method expects an array, you need to pass it an array, not a List:
Own[] arr = new Own[10]; // Or some size that makes sense...
in.readTypedArray(arr, CategoriesParceable.CREATOR);
There is a possibility to create an array filled with content of specified List. To achieve that you can call method toArray() of your list reference, for example:
Integer[] array = list.toArray(new Integer[list.size()]);
Suppose I have a class :
class Dummy{
public static ArrayList<String> varArray;
}
In another class I do this :
Class Dummy2{
void main()
{
ArrayList<String> temp = Dummy.varArray;
}
}
Now suppose in Dummy2 I add elements to temp. Will the changes be reflected in Dummy.varArray? Because this is what is happening in my program. I tried printing the address of the two and they both point to the same address. Didn't know static field worked like this. Or am I doing something wrong?
Its not about static. The statement ArrayList<String> temp = Dummy.varArray; means that both variables are referring to the same arraylist. As varArray is static, it will have only one copy.
You can read ArrayList<String> temp = Dummy.varArray; as, The variable temp is now referring to the ArrayList object which is being referred by Dummy.varArray
By the way, you need to initialize it using public static ArrayList<String> varArray = new ArrayList<String>(); before you perform any operations on it.
ArrayList<String> temp = Dummy.varArray; will take what is known as a reference copy (or a shallow copy). That is, they will point to the same object.
It does not take a deep copy. See How to clone ArrayList and also clone its contents?
Yes it is behaving correctly.
When you do this
ArrayList<String> temp = Dummy.varArray;
Both pointing to the same reference ,since temp not a new list, you just telling that refer to Dummy.varArray
To make them independent, create a new list
ArrayList<String> temp = new ArrayList<String>(); //new List
temp.addAll(Dummy.varArray); //get those value to my list
Point to note:
When you do this temp.addAll(Dummy.varArray) at that point what ever the elements in the varArray they add to temp.
ArrayList<String> temp = new ArrayList<String>(); //new List
temp.addAll(Dummy.varArray); //get those value to my list
Dummy.varArray.add("newItem");// "newitem" is not there in temp
The later added elements won't magically add to temp.
The static keyword means there will only be one instance of that variable and not one variable per instance.
What is the difference between
List<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia)); // Copy
List<Integer> list2 = Arrays.asList(ia);
, where ia is an array of integers?
I came to know that some operations are not allowed in list2. Why is it so?
How is it stored in memory (references / copy)?
When I shuffle the lists, list1 doesn't affect the original array, but list2 does. But still list2 is somewhat confusing.
How does ArrayList being upcasted to list differ from creating a new ArrayList?
list1 differs from (1)
ArrayList<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia));
First, let's see what this does:
Arrays.asList(ia)
It takes an array ia and creates a wrapper that implements List<Integer>, which makes the original array available as a list. Nothing is copied and all, only a single wrapper object is created. Operations on the list wrapper are propagated to the original array. This means that if you shuffle the list wrapper, the original array is shuffled as well, if you overwrite an element, it gets overwritten in the original array, etc. Of course, some List operations aren't allowed on the wrapper, like adding or removing elements from the list, you can only read or overwrite the elements.
Note that the list wrapper doesn't extend ArrayList - it's a different kind of object. ArrayLists have their own, internal array, in which they store their elements, and are able to resize the internal arrays etc. The wrapper doesn't have its own internal array, it only propagates operations to the array given to it.
On the other hand, if you subsequently create a new array as
new ArrayList<Integer>(Arrays.asList(ia))
then you create new ArrayList, which is a full, independent copy of the original one. Although here you create the wrapper using Arrays.asList as well, it is used only during the construction of the new ArrayList and is garbage-collected afterwards. The structure of this new ArrayList is completely independent of the original array. It contains the same elements (both the original array and this new ArrayList reference the same integers in memory), but it creates a new, internal array, that holds the references. So when you shuffle it, add, remove elements etc., the original array is unchanged.
Well, this is because ArrayList resulting from Arrays.asList() is not of the type java.util.ArrayList.
Arrays.asList() creates an ArrayList of type java.util.Arrays$ArrayList which does not extend java.util.ArrayList, but only extends java.util.AbstractList.
List<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia)); //copy
In this case, list1 is of type ArrayList.
List<Integer> list2 = Arrays.asList(ia);
Here, the list is returned as a List view, meaning it has only the methods attached to that interface. Hence why some methods are not allowed on list2.
ArrayList<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia));
Here, you are creating a new ArrayList. You're simply passing it a value in the constructor. This is not an example of casting. In casting, it might look more like this:
ArrayList list1 = (ArrayList)Arrays.asList(ia);
First of all, the Arrays class is a utility class which contains a number of utility methods to operate on Arrays (thanks to the Arrays class. Otherwise, we would have needed to create our own methods to act on Array objects)
asList() method:
asList method is one of the utility methods of Array class, it is a static method that's why we can call this method by its class name (like Arrays.asList(T...a) )
Now here is the twist. Please note that this method doesn't create new ArrayList object. It just returns a List reference to an existing Array object (so now after using asList method, two references to existing Array object gets created)
and this is the reason. All methods that operate on List object, may not work on this Array object using the List reference. Like
for example, Arrays size is fixed in length, hence you obviously can not add or remove elements from Array object using this List reference (like list.add(10) or list.remove(10);. Else it will throw UnsupportedOperationException).
any change you are doing using a list reference will be reflected in the exiting Arrays object (as you are operating on an existing Array object by using a list reference)
In the first case, you are creating a new Arraylist object (in the second case, only a reference to existing Array object is created, but not a new ArrayList object), so now there are two different objects. One is the Array object and another is the ArrayList object and there isn't any connection between them (so changes in one object will not be reflected/affected in another object (that is, in case 2, Array and Arraylist are two different objects)
Case 1:
Integer [] ia = {1,2,3,4};
System.out.println("Array : "+Arrays.toString(ia));
List<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia)); // new ArrayList object is created , no connection between existing Array Object
list1.add(5);
list1.add(6);
list1.remove(0);
list1.remove(0);
System.out.println("list1: " + list1);
System.out.println("Array: " + Arrays.toString(ia));
Case 2:
Integer [] ia = {1,2,3,4};
System.out.println("Array: " + Arrays.toString(ia));
List<Integer> list2 = Arrays.asList(ia); // Creates only a (new) List reference to the existing Array object (and NOT a new ArrayList Object)
// list2.add(5); // It will throw java.lang.UnsupportedOperationException - invalid operation (as Array size is fixed)
list2.set(0,10); // Making changes in the existing Array object using the List reference - valid
list2.set(1,11);
ia[2]=12; // Making changes in the existing Array object using the Array reference - valid
System.out.println("list2: " + list2);
System.out.println("Array: " + Arrays.toString(ia));
An explanation with documentation references would be better for someone looking for answer.
1. java.util.Arrays
This is a utility class with bunch of static methods to operate on given array
asList is one such static method that takes input array and returns an object of java.util.Arrays.ArrayList which is a static nested class that extends AbstractList<E> which in turn implements List interface.
So Arrays.asList(inarray) returns a List wrapper around the input array, but this wrapper is java.util.Arrays.ArrayList and not java.util.ArrayList and it refers to the same array, so adding more elements to the List wrapped array would affect the original one too and also we cannot change the length.
2. java.util.ArrayList
ArrayList has a bunch of overloaded constructors
public ArrayList() - // Returns arraylist with default capacity 10
public ArrayList(Collection<? extends E> c)
public ArrayList(int initialCapacity)
So when we pass the Arrays.asList returned object, i.e., List(AbstractList) to the second constructor above, it will create a new dynamic array (this array size increases as we add more elements than its capacity and also the new elements will not affect the original array) shallow copying the original array (shallow copy means it copies over the references only and does not create a new set of same objects as in original array)
String names[] = new String[]{"Avinash","Amol","John","Peter"};
java.util.List<String> namesList = Arrays.asList(names);
or
String names[] = new String[]{"Avinash","Amol","John","Peter"};
java.util.List<String> temp = Arrays.asList(names);
The above statement adds the wrapper on the input array. So the methods like add and remove will not be applicable on the list reference object 'namesList'.
If you try to add an element in the existing array/list then you will get "Exception in thread "main" java.lang.UnsupportedOperationException".
The above operation is readonly or viewonly.
We can not perform add or remove operation in list object.
But
String names[] = new String[]{"Avinash","Amol","John","Peter"};
java.util.ArrayList<String> list1 = new ArrayList<>(Arrays.asList(names));
or
String names[] = new String[]{"Avinash","Amol","John","Peter"};
java.util.List<String> listObject = Arrays.asList(names);
java.util.ArrayList<String> list1 = new ArrayList<>(listObject);
In the above statement you have created a concrete instance of an ArrayList class and passed a list as a parameter.
In this case, methods add and remove will work properly as both methods are from ArrayList class, so here we won't get any UnSupportedOperationException.
Changes made in the Arraylist object (method add or remove an element in/from an arraylist) will get not reflect in to the original java.util.List object.
String names[] = new String[] {
"Avinash",
"Amol",
"John",
"Peter"
};
java.util.List < String > listObject = Arrays.asList(names);
java.util.ArrayList < String > list1 = new ArrayList < > (listObject);
for (String string: list1) {
System.out.print(" " + string);
}
list1.add("Alex"); // Added without any exception
list1.remove("Avinash"); // Added without any exception will not make any changes in original list in this case temp object.
for (String string: list1) {
System.out.print(" " + string);
}
String existingNames[] = new String[] {
"Avinash",
"Amol",
"John",
"Peter"
};
java.util.List < String > namesList = Arrays.asList(names);
namesList.add("Bob"); // UnsupportedOperationException occur
namesList.remove("Avinash"); // UnsupportedOperationException
Note that, in Java 8, 'ia' above must be Integer[] and not int[]. Arrays.asList() of an int array returns a list with a single element. When using the OP's code snippet, the compiler will catch the issue, but some methods (e.g., Collections.shuffle()) will silently fail to do what you expect.
Many people have answered the mechanical details already, but it's worth noting:
This is a poor design choice, by Java.
Java's asList method is documented as "Returns a fixed-size list...". If you take its result and call (say) the .add method, it throws an UnsupportedOperationException. This is unintuitive behavior! If a method says it returns a List, the standard expectation is that it returns an object which supports the methods of interface List. A developer shouldn't have to memorize which of the umpteen util.List methods create Lists that don't actually support all the List methods.
If they had named the method asImmutableList, it would make sense. Or if they just had the method return an actual List (and copy the backing array), it would make sense. They decided to favor both runtime-performance and short names, at the expense of violating both the principle of least astonishment and the good object-oriented practice of avoiding UnsupportedOperationExceptions.
(Also, the designers might have made a interface ImmutableList, to avoid a plethora of UnsupportedOperationExceptions.)
package com.copy;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;
public class CopyArray {
public static void main(String[] args) {
List<Integer> list1, list2 = null;
Integer[] intarr = { 3, 4, 2, 1 };
list1 = new ArrayList<Integer>(Arrays.asList(intarr));
list1.add(30);
list2 = Arrays.asList(intarr);
// list2.add(40); Here, we can't modify the existing list,because it's a wrapper
System.out.println("List1");
Iterator<Integer> itr1 = list1.iterator();
while (itr1.hasNext()) {
System.out.println(itr1.next());
}
System.out.println("List2");
Iterator<Integer> itr2 = list2.iterator();
while (itr2.hasNext()) {
System.out.println(itr2.next());
}
}
}
Arrays.asList()
This method returns its own implementation of List. It takes an array as an argument and builds methods and attributes on top of it, since it is not copying any data from an array but using the original array this causes alteration in original array when you modify list returned by the Arrays.asList() method.
On the other hand, ArrayList(Arrays.asList());
is a constructor of ArrayList class which takes a list as argument and returns an ArrayList that is independent of list, i.e., Arrays.asList() in this case passed as an argument.
That is why you see these results.
1.List<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia)); //copy
2.List<Integer> list2 = Arrays.asList(ia);
In line 2, Arrays.asList(ia) returns a List reference of inner class object defined within Arrays, which is also called ArrayList but is private and only extends AbstractList. This means what returned from Arrays.asList(ia) is a class object different from what you get from new ArrayList<Integer>.
You cannot use some operations to line 2 because the inner private class within Arrays does not provide those methods.
Take a look at this link and see what you can do with the private inner class:
http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/util/Arrays.java#Arrays.ArrayList
Line 1 creates a new ArrayList object copying elements from what you get from line 2. So you can do whatever you want since java.util.ArrayList provides all those methods.
In response to some comments asking questions about the behaviour of Arrays.asList() since Java 8:
int[] arr1 = {1,2,3};
/*
Arrays are objects in Java, internally int[] will be represented by
an Integer Array object which when printed on console shall output
a pattern such as
[I#address for 1-dim int array,
[[I#address for 2-dim int array,
[[F#address for 2-dim float array etc.
*/
System.out.println(Arrays.asList(arr1));
/*
The line below results in Compile time error as Arrays.asList(int[] array)
returns List<int[]>. The returned list contains only one element
and that is the int[] {1,2,3}
*/
// List<Integer> list1 = Arrays.asList(arr1);
/*
Arrays.asList(arr1) is Arrays$ArrayList object whose only element is int[] array
so the line below prints [[I#...], where [I#... is the array object.
*/
System.out.println(Arrays.asList(arr1));
/*
This prints [I#..., the actual array object stored as single element
in the Arrays$ArrayList object.
*/
System.out.println(Arrays.asList(arr1).get(0));
// prints the contents of array [1,2,3]
System.out.println(Arrays.toString(Arrays.asList(arr1).get(0)));
Integer[] arr2 = {1,2,3};
/*
Arrays.asList(arr) is Arrays$ArrayList object which is
a wrapper list object containing three elements 1,2,3.
Technically, it is pointing to the original Integer[] array
*/
List<Integer> list2 = Arrays.asList(arr2);
// prints the contents of list [1,2,3]
System.out.println(list2);
Summary of the difference -
When a list is created without using the new, the operator Arrays.asList() method returns a wrapper which means:
you can perform an add/update operation.
the changes done in the original array will be reflected to List as well and vice versa.
Please consider my question. final values cannot be changed in java.
private final List<Integer> list = new ArrayList<Integer>();
above list instantiation is of final. now i can add any elements.
after that can i assign list=null?
Please help me.
Thanks!
That means the variable list is final.
Which means you can not assign something else to it again.
Once you are done assign a value (reference) to it as follows:
private final List<Integer> list = new ArrayList<Integer>();
You can not do something as below again:
list = new ArrayList<Integer>(); //this is invalid because you are trying to assign something else the variable list
list.add(new Integer(123)); is this code valid?
It's perfectly valid. You are just adding an object to the ArrayList that variable list is referencing.
The usage of final keyword is restricting the variable list not the ArrayList object that it's referencing.
The new keyword in Java creates a new Object and returns its reference. Hence, in your code, the list variable stores the reference to the new list. Declaring it final means that the reference stored in list is final, and cannot be changed.
The actual list is still modifiable.
For your code and if you have this:
private final List<Integer> list = new ArrayList<Integer>();
This is possible:
list.add(3);
This is not allowed:
list = new ArrayList<Integer>();
The final keyword indicates that a variable can only be initialized once. It does not guarantee immutability of the object assigned to that variable. In other words, it says something about what a variable can refer to, but nothing about the contents of the referent.
I have some data structures, and I would like to use one as a temporary, and another as not temporary.
ArrayList<Object> myObject = new ArrayList<Object>();
ArrayList<Object> myTempObject = new ArrayList<Object>();
//fill myTempObject here
....
//make myObject contain the same values as myTempObject
myObject = myTempObject;
//free up memory by clearing myTempObject
myTempObject.clear();
now the problem with this of course is that myObject is really just pointing to myTempObject, and so once myTempObject is cleared, so is myObject.
How do I retain the values from myTempObject in myObject using java?
You can use such trick:
myObject = new ArrayList<Object>(myTempObject);
or use
myObject = (ArrayList<Object>)myTempObject.clone();
You can get some information about clone() method here
But you should remember, that all these ways will give you a copy of your List, not all of its elements. So if you change one of the elements in your copied List, it will also be changed in your original List.
originalArrayList.addAll(copyArrayList);
Please Note: When using the addAll() method to copy, the contents of both the array lists (originalArrayList and copyArrayList) refer to the same objects or contents. So if you modify any one of them the other will also reflect the same change.
If you don't wan't this then you need to copy each element from the originalArrayList to the copyArrayList, like using a for or while loop.
There are no implicit copies made in java via the assignment operator. Variables contain a reference value (pointer) and when you use = you're only coping that value.
In order to preserve the contents of myTempObject you would need to make a copy of it.
This can be done by creating a new ArrayList using the constructor that takes another ArrayList:
ArrayList<Object> myObject = new ArrayList<Object>(myTempObject);
Edit: As Bohemian points out in the comments below, is this what you're asking? By doing the above, both ArrayLists (myTempObject and myObject) would contain references to the same objects. If you actually want a new list that contains new copies of the objects contained in myTempObject then you would need to make a copy of each individual object in the original ArrayList
Came across this while facing the same issue myself.
Saying arraylist1 = arraylist2 sets them both to point at the same place so if you alter either the data alters and thus both lists always stay the same.
To copy values into an independent list I just used foreach to copy the contents:
ArrayList list1 = new ArrayList();
ArrayList list2 = new ArrayList();
fill list1 in whatever way you currently are.
foreach(<type> obj in list1)
{
list2.Add(obj);
}
Supopose you want to copy oldList into a new ArrayList object called newList
ArrayList<Object> newList = new ArrayList<>() ;
for (int i = 0 ; i<oldList.size();i++){
newList.add(oldList.get(i)) ;
}
These two lists are indepedant, changes to one are not reflected to the other one.
Lets try the example
ArrayList<String> firstArrayList = new ArrayList<>();
firstArrayList.add("One");
firstArrayList.add("Two");
firstArrayList.add("Three");
firstArrayList.add("Four");
firstArrayList.add("Five");
firstArrayList.add("Six");
//copy array list content into another array list
ArrayList<String> secondArrayList=new ArrayList<>();
secondArrayList.addAll(firstArrayList);
//print all the content of array list
Iterator itr = secondArrayList.iterator();
while (itr.hasNext()) {
System.out.println(itr.next());
}
In print output as below
One
Two
Three
Four
Five
Six
We can also do by using clone() method for which is used to create exact copy
for that try you can try as like
**ArrayList<String>secondArrayList = (ArrayList<String>) firstArrayList.clone();**
And then print by using iterator
**Iterator itr = secondArrayList.iterator();
while (itr.hasNext()) {
System.out.println(itr.next());
}**
You need to clone() the individual object. Constructor and other methods perform shallow copy. You may try Collections.copy method.
Straightforward way to make deep copy of original list is to add all element from one list to another list.
ArrayList<Object> originalList = new ArrayList<Object>();
ArrayList<Object> duplicateList = new ArrayList<Object>();
for(Object o : originalList) {
duplicateList.add(o);
}
Now If you make any changes to originalList it will not impact duplicateList.
to copy one list into the other list, u can use the method called
Collection.copy(myObject myTempObject).now after executing these line of code u can see all the list values in the myObject.
Copy of one list into second is quite simple , you can do that as below:-
ArrayList<List1> list1= new ArrayList<>();
ArrayList<List1> list2= new ArrayList<>();
//this will your copy your list1 into list2
list2.addAll(list1);
Here is a workaround to copy all the objects from one arrayList to another:
ArrayList<Object> myObject = new ArrayList<Object>();
ArrayList<Object> myTempObject = new ArrayList<Object>();
myObject.addAll(myTempObject.subList(0, myTempObject.size()));
subList is intended to return a List with a range of data. so you can copy the whole arrayList or part of it.
Suppose you have two arraylist of String type .
Like
ArrayList<String> firstArrayList ;//This array list is not having any data.
ArrayList<String> secondArrayList = new ArrayList<>();//Having some data.
Now we have to copy the data of second array to first arraylist like this,
firstArrayList = new ArrayList<>(secondArrayList );
Done!!
The simplest way is:
ArrayList<Object> myObject = new ArrayList<Object>();
// fill up data here
ArrayList<Object> myTempObject = new ArrayList(myObject);