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Usage of Iteration's .hasNext(); and .next(); methods in Java

For 2 days I'm pretty confused about .hasNext(); and next(); methods of Iteration interface especially in while loop. Let me give an example:
import java.util.*; // imported whole java.util package.
class Main {
public static void main(String[] args) {
ArrayList<String> cars = new ArrayList<String>(); // created ArrayList which name is cars.
cars.add("Volvo");
cars.add("Mercedes");
cars.add("BMW");
Iterator<String> x = cars.iterator();
while(x.hasNext()) {
System.out.print(x.next() + " "); // It prints Volvo Mercedes BMW
}
}
}
I understood that .hasNext(); is boolean and it returns true if iteration has elements. The .next(); method returns the next element. After first element Volvo, it gets back while(x.hasNext()) and entering the inside of loop again but where is the counter of this loop? I mean after printing Volvo how can it goes to the next element? It returns all element and if there is no more .hasNext(); returns false and code continues to next line is simple answer and correct but I want to understand it clearly.

Actually the iterator() Method Creates an iterator for all elements in your (let's say) Arraylist. Now in Your Code, the condition of while loop x.hasNext() checks whether the list contains an element or not, if yes it will return true else false.
Now x.next() point to First Element (Like of LinkedLists for example) in Your ArrayList and store that Object(Volvo in your case). When You Call this method it basically gives you reference of that object in List and the Iterator Moves to next element in the List. When Calling next() method the iterator(x in your case) returns the object and then moves to next Element (Mercedes in your case).
Now When you call next() method Again you will find Mercedes is returned. To know How it Works type System.out.println(x.next()) thrice instead of while loop u will understand that it moves to next location. If you type System.out.println(x.next()) fourth time it will give exception because there is no element in your list furthur more. Exception in thread "main" java.util.NoSuchElementException This is the Exception.
That's why hasNext() method is used as it checks whether an element is there or not.
you can Compare this to printing of linkedlist(Data structure if you know) where we make one object point to head node print it and move to next node. Same Case is here it returns current element(object) and moves to next element(object).

while is java-ese for: Keep doing this until a thing changes.
It's a bit like this common household task:
How to wash dishes
Check if there are still dirty dishes on the right side of the counter.
If there are some, do the dishwash thing: Pick up the nearest dirty dish, and wash it, then stow it away on the left side of the counter, and go back to the start of this algorithm.
Otherwise (there are no dirty dishes), you are done.
while (counterRightSide.hasItems()) {
Dish dirty = counterRightSide.fetch();
Dish clean = sink.clean(dirty);
counterLeftSide.stow(clean);
}
EDIT: I realize now that 'a kitchen counter' is an unfortunate example, given the homonym 'kitchen counter' and 'counter in code'. I'll use 'accumulator' instead of 'counter in code' to fix this confusion.
Note how there is no accumulator here either, and you aren't counting in your head when you wash the dishes either. You COULD first take inventory of the amount of dirty dishes you have (say, 15 dishes), and then write a protocol where you grab a dirty dish exactly 15 times before decreeing that you've done the job, but surely you realize that that's just one way to do it, and another is to just... check if there are any dirty dishes left every time you're done washing 1 dirty dish. Which is how the above code works.
Note that the action of 'fetching' an item from the right side of the kitchen counter changes properties of that kitchen counter. It now has one less item on it.
The same applies to iterators: Calling .next() on an iterator changes it in some form. After all, if you call next twice, the results of these calls are different. Contrast with invoking, say, someArrayList.get(5) a bunch of times in a row; that doesn't change the arraylist at all, and therefore you get the same thing back every time. next() is not like that.
So how DOES that work? Who counts this stuff?
Well, that's the neat thing about abstraction! A 'Collection' is something that can do a few things, such as 'it must be able to report to how how many items it holds', and, 'it must be able to festoon you an object that can be used to loop through each item that it holds'.
That's what an iterator is: An object that can do that.
So how does it work? Who knows! It doesn't matter! As long as you CAN do these things, you can be a collection, you are free to implement the task however you like.
Okay, but how does a common collection, like, say, ArrayList do this?
With a counter, of course. Here's the implementation:
public Iterator<T> iterator() {
return new Iterator<T>() {
int counter = 0; // here it is!
public boolean hasNext() {
return counter < size();
}
public int next() {
return get(counter++);
}
};
}
you're getting this object and referencing it in your code (your x variable), and that object has a field with a counter in it. You can't see it; it's private, and not actually in the Iterator type (Iterator is an interface; what you get is some unknown subtype of it, and that contains the counter variable). That's assuming you're getting an iterator by invoking .iterator() on an arraylist. If you invoke it on something else, there may or may not be a counter - as long as the iterator WORKS, it doesn't matter how it works, that's the beauty of interfaces.

A while loop checks the condition and when condition is true then executes the body of the loop and iterates itself,
while(condition){
//do something.
}
The hasNext() is a method from the Iterator interface which returns true if the "iteration" has more elements, if there are no more elements it returns fals and will no longer enter the body of the while loop:
while(x.hasNext){
//do something.
}
The next() method is a method from the Iterator interface and returns the next element of the iteration.
while(x.hasNext){
x.next();
}

I mean after printing Volvo how can it goes to the next element? It
returns all element and if there is no more .hasNext(); returns false
and code continues to next line is simple answer and correct but I
want to understand it clearly.
int cursor; // index of next element to return
//...
public boolean hasNext() {
return cursor != size;
}
public E next() {
checkForComodification();
int i = cursor;
if (i >= size)
throw new NoSuchElementException();
Object[] elementData = ArrayList.this.elementData;
if (i >= elementData.length)
throw new ConcurrentModificationException();
cursor = i + 1;
return (E) elementData[lastRet = i];
}
//...
Given above is how ArrayList#hasNext and and ArrayList#next have been implemented. The code is straight forward and easy to understand. You can check the full code by using a decompiler or here.

I can't get to modify my static variable in java

You give a grid (4x4 here). you need to find out the total no of unique paths from (0,0) to (4,4). main() call a function pathify for this. It finds the possible "next steps" and calls it again. When (4,4) is reached noOfPaths++; is supposed to execute. This doesn't happen and I can't find the problem.
import java.util.ArrayList;
public class NoOfPaths {
static int xRows = 4;
static int yColumns = 4;
static int noOfPaths = 0;
/*A robot is located in the upper-left corner of a 4×4 grid.
* The robot can move either up, down, left, or right,
* but cannot go to the same location twice.
* The robot is trying to reach the lower-right corner of the grid.
* Your task is to find out the number of unique ways to reach the destination.
**/
static ArrayList validNeighbours (int x,int y, ArrayList visited) {
ArrayList valid = new ArrayList();
if((x+1 <= xRows) && !visited.contains(((x+1)*10)+y) ) {
valid.add(((x+1)*10)+y);
}
if((x-1 >= 0) && !visited.contains(((x-1)*10)+y) ) {
valid.add(((x-1)*10)+y);
}
if((y+1 <= yColumns) && !visited.contains(x*10+y+1) ) {
valid.add(x*10+y+1);
}
if((y-1 >= 0) && !visited.contains(x*10+y-1) ) {
valid.add(x*10+y-1);
}
return valid;
}
static void pathify(int x,int y, ArrayList alreadyVisited) {
if(x == xRows && y == yColumns) {
noOfPaths++;
} else {
alreadyVisited.add(x*10+y);
ArrayList callAgain = new ArrayList();
callAgain = validNeighbours(x,y,alreadyVisited);
for (int t=0,temp; t<callAgain.size(); t++) {
temp=(int) callAgain.get(t);
pathify(temp/10, temp%10, alreadyVisited);
}
}
}
public static void main(String[] args) {
ArrayList alreadyVisited = new ArrayList();
pathify(0, 0, alreadyVisited);
System.out.println(noOfPaths);
}
}

The error is in how you're handling alreadyVisited. The first time pathify is called, this list will contain only the initial square (0,0), which is fine. Here's the important part of your code:
for (int t=0,temp; t<callAgain.size(); t++) {
temp=(int) callAgain.get(t);
pathify(temp/10, temp%10, alreadyVisited);
}
You've found the neighbors of the initial cell. Your code will pick the first neighbor; then it will find paths starting with that neighbor, and the recursive calls to pathify will add cells to alreadyVisited.
Now, after all the recursive calls come back, you're ready to find cells starting with the second neighbor of the initial cell. But you have a problem: alreadyVisited still has all the cells it's collected from the paths it found starting with the second neighbor. So you won't find all possible paths starting with the second neighbor; you won't find any path that includes any cell in any path you've previously found. This isn't what you want, since you only want to avoid visiting the same cell in each path--you don't want to avoid visiting the same cell in all your previous paths. (I simplified this a little bit. In reality, the problem will start occurring deeper down the recursive stack, and you won't even find all the paths beginning with the first neighbor.)
When implementing a recursive algorithm, I've found that it's generally a bad idea to keep an intermediate data structure that is shared by recursive invocations that will be modified by those invocations. In this case, that's the list alreadyVisited. The problem is that when an invocation deeper down the stack modifies the structure, this affects invocations further up, because they will see the modifications after the deeper invocations return, which is basically data they need changing underneath them. (I'm not talking about a collection that is used to hold a list of results, if the list is basically write-only.) The way to avoid it here is that instead of adding to alreadyVisited, you could create a clone of this list and then add to it. That way, a deeper invocation can be sure that it's not impacting the shallower invocations by changing their data. That is, instead of
alreadyVisited.add(x*10+y);
write
alreadyVisited = [make a copy of alreadyVisited];
alreadyVisited.add(x*10+y);
The add will modify a new list, not the list that other invocations are using. (Personally, I'd declare a new variable such as newAlreadyVisited, since I don't really like modifying parameters, for readability reasons.)
This may seem inefficient. It will definitely use more memory (although the memory should be garbage-collectible pretty quickly). But trying to share a data structure between recursive invocations is very, very difficult to do correctly. It can be done if you're very careful about cleaning up the changes and restoring the structure to what it was when the method began. That might be necessary if the structure is something like a large tree, making it unfeasible to copy for every invocation. But it can take a lot of skill to make things work.
EDIT: I tested it and it appears to work: 12 if xRows=yColumns=2, 8512 if both are 4 (is that correct?). Another approach: instead of copying the list, I tried
alreadyVisited.remove((Object)(x*10+y));
at the end of the method ((Object) is needed so that Java doesn't think you're removing at an index) and that gave me the same results. If you do that, you'll make sure that alreadyVisited is the same when pathify returns as it was when it started. But I want to emphasize that I don't recommend this "cleanup" approach unless you really know what you're doing.

How to Determine If a Collection Is Empty

I have a 3 dimensional ArrayList and I want to determine if it is empty or not. There is an exception called EmptyCollectionException which is not part of java standard library and hence I'm not allowed to use it.
Is there a way to accomplish that using a native java exception or function?
The 3D List is constructed as follow:
public void makeRandomCardListForLearning (Course courseToBeMadeRandom) {
List<List<List<Card>>> course = new ArrayList<List<List<Card>>>();
for(Chapter chptr: courseToBeMadeRandom.getChapterList()) {
List<List<Card>> chapter = new ArrayList<List<Card>>();
course.add(chapter);
for(Deck dck: chptr.getDeckList()) {
List<Card> deck = new LinkedList<Card>();
chapter.add(deck);
for(Card crd: dck.getCardList()) {
if(dck.isTheCardDueToday(crd.getLastDate())) {
deck.add(crd);
}
}
Collections.shuffle(deck);
}
}
}
As I go through course, chapter and deck I create a List for each one. There is only one course, many chapters, many decks and of course many cards which are saved under deck doublyLinkedList if they pass the pre-condition. If no card passes the condition, I have a 3D list which exists but has no cards. And I want to determine that If no card exists in the list, then the user receives an error message.
In fact I only need the cards. But I also need to know in which deck each card resides at the moment. If I just make a list and go through all chapters and decks and put cards in that list based on the condition then I have no clue in which chapter and deck each card resides. That can be solved by maybe adding two attributes to the card class. But that was a mistake as we designed the system and adding them now costs a lot of change in other parts of the program. Each index in course List represents the chapter number and each index in chapter list represents the deck number. I solved the problem that way.

This should do it:
public static boolean isEmpty(List<List<List<Card>>> list3d) {
return list3d.stream().flatMap(llc -> llc.stream()).flatMap(lc -> lc.stream()).count() == 0;
}
It takes into account that the outer lists may contain empty inner lists. It deems the entire 3D list empty if there are no cards in it.

you can do something like this
List<List<List<Card>>> course = new ArrayList<List<List<Card>>>();
// some possible codes
boolean check = course.isEmpty()
// other possible codes
if (check) {
// do something
}
or any arraylist you want to check or any way you want to reach your goal

remotely remove item from arraylist after intersect

I am working on a game project. So far so good, but i just stuck on ome basic thing and i cant find a solution and make it work properly. I decided to come here and ask you ppl of suggestions.
PROBLEM:
When the player comes to contact with a diamond, i suppose to remove the diamond from the level and from the arraylist containing all the objects in the world. What always happens i get an exception error message after remove() method called.
CODES:
1.Class with the list: EDIT_1
private ArrayList<AbstractObject> objects = new ArrayList<AbstractObject>();
public void removeObject(String name){
ArrayList<AbstractObject> newest = new ArrayList<AbstractObject>();
ListIterator<AbstractObject> delete=objects.listIterator();
while(delete.hasNext()){
if(name.equals(delete.next().getName())){
delete.remove();
}
else{
delete.previous();
newest.add(delete.next());
}
}
objects=newest;
}
2.Player class calling the removeObject method: EDIT_1
public void playerLogic(){
fallingDown();
for(AbstractObject object : this.getWorld().getListOfObjects()){ <--------ERROR HERE
if(this.intersects(object)){
if(object instanceof FinishZone && points>=getWorld().getDiamondCount()){
if(!(getWorld().getManager().isMoreLevels())){
getWorld().getMenu().openMenu(true);
}
else{
this.getWorld().getManager().nextLevel();
}
}
if(object instanceof Diamond){
points++;
this.getWorld().removeObject(object.getName());
}
}
}
}
ERROR:
Exception in thread "Thread-2" java.util.ConcurrentModificationException
at java.util.ArrayList$Itr.checkForComodification(ArrayList.java:859)
at java.util.ArrayList$Itr.next(ArrayList.java:831)
at project.objects.characters.Player.playerLogic(Player.java:77)
at project.objects.characters.Player.update(Player.java:70)
at project.world.World.update(World.java:110)
at project.Main.update(Main.java:122)
at project.Main.run(Main.java:65)
at java.lang.Thread.run(Thread.java:745)
I checked up some examples of removing items from arraylist but i havent find the difference.
EDIT_1:
So i figured out how to do it but i always get the error. I edited the removeobject code block. This worked good with a neutral list that i created for testing. I put all the items which i dont want to delete into a new list than ovewritten the old arraylist with the newest one. It worked with no exception error. When i made the same with the game list i want to edit it thrown the same error.
Ill put there the render code too if maybe there is the problem...
public void render(Graphics g) {
if(menu.getChoice()==-1){
menu.render(g);
}
else if(menu.getChoice()==0){
g.setColor(Color.white);
for(AbstractObject tempObj : objects){
tempObj.render(g);
}
}
}
FIXED:
Ill changed the starting list is ListIterator instead of putting items in arrayList before adding it to ListIterator. All methods changed to iterate. Working fine :)

You can't remove object while iterating over a list.
One option - use iterator.remove() - if you iterate with iterator, not the "enhanced for loop". You'll need to slightly modify your loop code, but the functionality will be the same.
Another: Store all objects to remove in an auxiliary list, and remove them all at the end of the loop.

Removing Actors does not delet all Actors

I am currently trying to save special Actors so i can put them on a map again if the old map get loaded. Therefor i want to put them into a HashMap<String, ArrayList<Monster>> monsterAtMap and remove them from there Stages. So i am trying this:
private void saveMonsters() {
if (this.screen.figureStage.getActors().size == 0)
return;
ArrayList<Monster> monsters = new ArrayList<Monster>();
for (Actor a : this.screen.figureStage.getActors()) {
a.remove();
}
Gdx.app.log("Figurstage size", ""+ this.screen.figureStage.getActors().size);
this.monsterAtMap.put(this.currentMap.name, monsters);
}
As start. But i noticed that it does not delete all. It does just delete 10 thats all. I do log the size of it befor and after the deleting. It's current 21 (20Monsters and 1 Character) after delete the size is 11.I also added this this.screen.figureStage.getRoot().removeActor(a); but this does not change anything.
Any Idea to that?
[EDIT] I wrote a workaround so my idea is working but the general idea that should work isnt possible because the .remove() does not always delete the Actor in anyway?! The workaround does look like this:
private void saveMonsters() {
this.chara = this.screen.character;
if (this.screen.figureStage.getActors().size == 0)
return;
ArrayList<Monster> monsters = new ArrayList<Monster>();
for (Actor a : this.screen.figureStage.getActors()) {
if (a.getClass() == Monster.class)
monsters.add((Monster) a);
}
this.screen.figureStage.clear();
this.screen.figureStage.addActor(chara);
this.monsterAtMap.put(this.currentMap.name, monsters);
}
The .clear()does work correct.

Deleting objects from a container while iterating over that container is always fraught with issues and complications, and I think you're running into some of these issues with the Stage's list of actors. The Stage code tries to use SnapshotArray to hide some of these issues, but its not clear to me that it will work with the code you've written.
One way to avoid this would be to loop through getActors() once and copy the actors into the monsters array, then loop through the monsters array and remove the actors from the Stage (or invoke figureStage.getRoot().clearChildren()). This should prevent you from iterating over a list that you're modifying.
Alternatively, look at how Group.clearChildren() is implemented (it uses an explicit integer index in the array of children, and not an iterator over the Array, and avoid some of the issues).