Weird Java Concurrent modification exception example [duplicate] - java

This question already has answers here:
Why am I not getting a java.util.ConcurrentModificationException in this example?
(10 answers)
Closed 4 years ago.
If we write like this, there is a concurrent modification exception :
public static void main(String... args) {
List<String> listOfBooks = new ArrayList<>();
listOfBooks.add("Programming Pearls");
listOfBooks.add("Clean Code");
listOfBooks.add("Effective Java");
listOfBooks.add("Code Complete");
System.err.println("Before deleting : " + listOfBooks);
for (String book : listOfBooks) {
if (book.contains("Code")) {
listOfBooks.remove(book);
}
}
System.err.println("After deleting : " + listOfBooks);
}
On the other hand, if we write like this, there is NO concurrent modification exception !
Notice that code is exact the same, except the strings for compare, in first example it is a Code, and in second it is a Java
public static void main(String... args) {
List<String> listOfBooks = new ArrayList<>();
listOfBooks.add("Programming Pearls");
listOfBooks.add("Clean Code");
listOfBooks.add("Effective Java");
listOfBooks.add("Code Complete");
System.err.println("Before deleting : " + listOfBooks);
for (String book : listOfBooks) {
if (book.contains("Java")) {
listOfBooks.remove(book);
}
}
System.err.println("After deleting : " + listOfBooks);
}
I'm using Netbeans 8.2, Windows 7 32bit, with JDK 1.8.0_131
What's wrong ?

List.remove() will not throw ConcurrentModificationException when it removes the second last element from the list.
Quoting from this Java Bug (JDK-4902078) .
When the Collections Framework was added to the platform it was deemed too expensive to check for comodification once rather than twice per iteration; the check was made on Iterator.next rather than Iterator.hasNext. Expert reviewers thought this was sufficient. They were unaware that it fails to detect one important case: if an element is removed from the list immediately prior to the final call to hasNext in an iteration, the call returns false and the iteration terminates, silently ignoring the last element on the list.
You can also check this answer :-
https://stackoverflow.com/a/8189786/1992276

There are two ways used to iterate over an collection: enumeration and iterator.
First one allows for the collection to be modified during iteration (fail slow), second does not (fail fast). In a for-each loop you are using an iterator, so any modification to the collection, during it's iteration would cause an exception.
You have 3 choices, to solve this problem:
Use an iterator instead:
Iterator<String> bookIt = listOfBooks.iterator();
while(bookIt.hasNext()){
String book = bookIt.next();
if (book.contains("Java")) {
bookIt.remove();
}
}
Create a new list with only acceptable elements (filter out the unwanted):
List<String> booksWithNoCode = listOfBooks.stream()
.filter(book-> !book.contains("Code"))
.collect(toList())
Use Collection.removeIf(), you will remove all elements from the list, that are matching given criteria.
listOfBooks.removeIf(book-> book.contains("Code"))
You can find more information in this post and here.

You can't modify the listOfBooks while you are iterating though it with the for each loop.
edit:
for (String book : listOfBooks) {
if (book.contains("Code")) {
listOfBooks.remove(book);
}
}
Is the same as:
for (Iterator<String> i = listOfBooks.iterator(); i.hasNext();) {
String book = i.next();
if (book.contains("Code")) {
listOfBooks.remove(book);
}
}
http://hg.openjdk.java.net/jdk8/jdk8/jdk/file/tip/src/share/classes/java/util/ArrayList.java
The key in the arraylist code is:
public boolean remove(Object o) {
if (o == null) {
for (int index = 0; index < size; index++)
if (elementData[index] == null) {
fastRemove(index);
return true;
}
} else {
for (int index = 0; index < size; index++)
if (o.equals(elementData[index])) {
fastRemove(index);
return true;
}
}
return false;
}
/*
* Private remove method that skips bounds checking and does not
* return the value removed.
*/
private void fastRemove(int index) {
modCount++;
int numMoved = size - index - 1;
if (numMoved > 0)
System.arraycopy(elementData, index+1, elementData, index,
numMoved);
elementData[--size] = null; // clear to let GC do its work
}
and the iterator code:
public boolean hasNext() {
return cursor != size;
}
#SuppressWarnings("unchecked")
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];
}
The cursor always points to the next element so when you get the "Effective Java" i = 2 but cursor is 3.
When you call the remove the cursor is at 3 and the size is 4.
The size is then decremented by the remove and now cursor == size and the next hasNext() returns false ending the loop.

Related

String list get an item starting without loop

I have a array list contains thousands of data.
For Example:
List<String> custNames = new ArrayList<String>();
custNames.add("John");
custNames.add("Tom");
custNames.add("Bart");
custNames.add("Tim");
custNames.add("Broad");
Now I want to get count of names only starting with 'T'. I used looping mechanism for my solution.
List<String> filterNames = new ArrayList<String>();
String nameStarts="T";
for(int i=0;i<custNames.size();i++)
{
if(custNames.get(i).toLowerCase().startsWith(nameStarts.toLowerCase()))
{
filterNames.add(custNames.get(i));
}
}
System.out.println(filterNames.size());
But I have very large collection of data in this custNames list.
Is there any different solution without using loop?
Thanks.
There is very good solution from Java 8 for your problem.
Try this,
long filterNameCount = custNames
.stream()
.parallel()
.filter((s) -> s.startsWith(nameStarts.toLowerCase()))
.count();
System.out.println(filterNameCount);
If you are open to using a third-party library, there are a few interesting options you could use with Eclipse Collections.
If you use the ArrayList as you have it above, you can use the LazyIterate utility as follows:
int count = LazyIterate.collect(custNames, String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
Assert.assertEquals(2, count);
If you use the Eclipse Collections replacement for ArrayList, you can use the rich functional protocols available directly on MutableList:
MutableList<String> custNames =
Lists.mutable.with("John", "Tom", "Bart", "Tim", "Broad");
String nameStarts= "T";
int count = custNames.asLazy()
.collect(String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
System.out.println(count);
Assert.assertEquals(2, count);
The serial API in Eclipse Collections is eager-by-default, which is why I called asLazy() first. The collect method would otherwise create another MutableList.
If you benchmark your code with your full set of data, the following parallel version of the code may be more performant:
MutableList<String> custNames =
Lists.mutable.with("John", "Tom", "Bart", "Tim", "Broad");
String nameStarts= "T";
int processors = Runtime.getRuntime().availableProcessors();
int batchSize = Math.max(1, custNames.size() / processors);
ExecutorService executor = Executors.newFixedThreadPool(processors);
int count = custNames.asParallel(executor, batchSize)
.collect(String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
executor.shutdown();
Assert.assertEquals(2, count);
The asParallel() API in Eclipse Collections is lazy-by-default. The API forces you to pass in a an ExecutorService and an int batchSize. This gives you complete control over the parallelism.
You can also use the Stream API with all MutableCollections in Eclipse Collections because they extend java.util.Collection.
Note: I am a committer for Eclipse Collections.
You could also use a tree storage : it would very efficient for this kind of search. If you are stucked with a list the previous answered is a way to do.
remove all the items which dont start with "T" like this:
custNames.removeIf(p->!p.startsWith("T"));
you can make a copy out of your list and remove items not starting with "T".
First, you can shorten your initialization with Arrays.asList(T); Second, I would use a simple loop to build a table of counts once and then use that to determine the subsequent queries. Something like,
List<String> custNames = new ArrayList<String>(Arrays.asList("John", "Tom",
"Bart", "Tim", "Broad"));
int[] counts = new int[26];
for (String name : custNames) {
char ch = Character.toLowerCase(name.charAt(0));
counts[ch - 'a']++;
}
for (int i = 0; i < counts.length; i++) {
if (counts[i] > 0) {
System.out.printf("There are %d words that start with %c%n",
counts[i], (char) ('a' + i));
}
}
Which outputs
There are 2 words that start with b
There are 1 words that start with j
There are 2 words that start with t
Or, in the specific case - counts['t' - 'a'] is the count of words starting with t.
If you have more or less static list and perform search operation often you can sort your list or use TreeMap.
Also you don't need to create new list and get its size then. You can simply create a counter variable and increment it.
You can create your own sorting and finding implementation.
Consider the following:
public class ContainingArrayList<E> extends ArrayList<E> {
private Comparator<E> comparator;
public ContainingArrayList(Comparator<E> comparator) {
this.setComparator(comparator);
}
#Override
public boolean add(E e) {
// If the collection is empty or the new element is bigger than the last one, append it to the end of the collection
if(size() == 0 || comparator.compare(e, get(size()-1)) >= 0)
return super.add(e);
else {
for (int i = 0; i < size(); i++) {
int result = comparator.compare(e, get(i));
// If the new element is bigger than the current element, continue with the next element
if (result > 0) continue;
// If the new element is equal to the current element, no need to insert (you might insert of course)
if (result == 0) return false;
// Otherwise the new element is smaller than the current element, so insert it between the previous and the current element
super.add(i, e);
return true;
}
return super.add(e);
}
}
public E get(E containingElement) {
int start = 0;
int end = size()-1;
// If the element is the first one, return the first element
if(comparator.compare(containingElement, super.get(start)) == 0)
return super.get(start);
// If the element is the last one, return the last element
if(comparator.compare(containingElement, super.get(end)) == 0)
return super.get(end);
// Otherwise do a binary search
while(start != end) {
// Get the element between start and end positions
E mid = super.get(start + (end/2));
// Compare the two elements
int result = comparator.compare(containingElement, mid);
// If the middle element compared to the containing element is equal, return the middle element
if(result == 0) {
return mid;
}
// If the containing element is smaller than the middle, halve the end position
else if(result < 0) {
end = start + (end/2);
}
// If the containing element is bigger than the middle, set the start position to the middle position
else if(result > 0) {
start = start + (end/2);
}
}
return null;
}
public Comparator<E> getComparator() {
return comparator;
}
public void setComparator(Comparator<E> comparator) {
this.comparator = comparator;
}
}
The custom comparator is used to sort the elements and to find the element that starts with a specific character. This means that you can change the comparator implementation for your needs at any time or you can create a more dynamic finding solution.
Test:
public class SortFindTest {
public SortFindTest() {
ContainingArrayList<String> t = new ContainingArrayList<String>(new MyComparator());
t.add("John");
t.add("Tom");
t.add("Bart");
t.add("Tim");
t.add("Broad");
System.out.println(t.get("T"));
}
class MyComparator implements Comparator<String> {
#Override
public int compare(String o1, String o2) {
int o1c = o1.charAt(0);
int o2c = o2.charAt(0);
if(o1c == o2c)
return 0;
if(o1c > o2c)
return 1;
return -1;
}
}
public static void main(String[] args) {
new SortFindTest();
}
}
I'm not sure if this would be faster than Java 8 Stream API but it worth a try.
If the order in which the items are stored does not matter, you could store the names in a HashMap, where the first character of each name is the key, and an ArrayList of names with that first character are the values. And then all you need to do, assuming the HashMap is named customerList, is customerList.get("T").size().
Initializing HashList and Adding Customers
HashMap<Character, ArrayList<String>> customerList = new HashMap<Character, ArrayList<String>>();
int NUM_ALPHABETS = 26;
int ascii_char = 97;
for(int i = 0; i < NUM_ALPHABETS; i++){
char c = (char) ascii_char;
customerList.add(c, new ArrayList<String>());
ascii_char++;
}
customerList.get("t").add("Tony");
customerList.get("a").add("Alice");
customerList.get("b").add("Ben");
Getting Number of Customers Starting with "t"
int num_t = customerList.get("t").size();

Find ArrayList method to Find Array method

I made a method which finds a value in my ArrayList. I also copied this method so I could use it for my array but certain things such as the get and size don't work. I'm unsure how I'm supposed to restructure it.
public Product findProduct(String givenProduct) throws IllegalProductCodeException {
IllegalProductCodeException notFoundMessage
= new IllegalProductCodeException("Product was not found");
int size = rangeOfProducts.length;
int i = 0;
boolean productFound = false;
while (!productFound && i < size) { //While book hasn't been found and i is less than the size of the array
productFound = rangeOfProducts.get(i).getProductCode().equals(givenProduct);
//Checks whether the given value in the array's reference is equal to the given reference entered
i++; //if not then add 1
}
if (productFound) {
return rangeOfProducts.get(i - 1);
} else {
throw notFoundMessage;
}
}
The array alternative to .get(i) will be [i] and the alternative to .size() will be .length.
for (Product product : products) {
if (product.getProductCode().equals(givenProduct)) {
return product;
}
}
throw new IllegalProductCodeException("Product was not found");
Edit: Java 5's enhanced for loop is equivalent to
for (Iterator<Product> iter=products.iterator(); iter.hasNext(); ) {
Product product = iter.next();

I have to return a value from a function that is called multiple times and I am unable to do that

I am stuck.
The following function is supposed to return currVm, an integer. But if I make a return I will break the loop and next time when this function is called,the same process will begin again.
What shall I do, so that I continue from where I left off ? I tried making static variables but I that didn't help me.
#Override
public int getNextAvailableVm() {
Set<String> dataCenters = confMap.keySet();
for (String dataCenter : dataCenters) {
LinkedList<DepConfAttr> list = confMap.get(dataCenter);
Collections.sort(list, new MemoryComparator());
int size = list.size() - 1;
int count = 0;
while(size >= 0) {
DepConfAttr dca = (DepConfAttr)list.get(count);
int currVm = dca.getVmCount();
int c = 0;
while(c <= currVm) {
allocatedVm(currVm);
c++;
return currVm;
}
count++;
size--;
}
}
return 0;
}
The for-each loop assigns a new data center that acts as a key for the confMap.The list that I get as a value, is sorted.Then a loop is run till it escapes its size.Inside this while loop, another while loop is run from where a function named allocatedVm of the inherited class is called. A parameter named currVm is passed to it.
This is the variable that I need to return. What shall I do to return this variable ? I have to start from I left off. I mean the next call should appear to be the next step, whatever it was, while executing the loop.
Add List<Integer> object to your class, and change your method as follows:
private Iterator<Integer> availableVms = null;
#Override
public int getNextAvailableVm() {
if (availableVms != null) {
if (availableVms.hasNext()) {
return availableVms.next();
}
return 0;
}
List<Integer> tmp = new ArrayList<Integer>();
Set<String> dataCenters = confMap.keySet();
for (String dataCenter : dataCenters) {
LinkedList<DepConfAttr> list = confMap.get(dataCenter);
Collections.sort(list, new MemoryComparator());
int size = list.size() - 1;
int count = 0;
while(size >= 0) {
DepConfAttr dca = (DepConfAttr)list.get(count);
int currVm = dca.getVmCount();
int c = 0;
while(c <= currVm) {
allocatedVm(currVm);
c++;
tmp.add(currVm);
}
count++;
size--;
}
}
availableVms = tmp.iterator();
return availableVms.hasNext() ? availableVms.next() : 0;
}
The idea is to pre-generate the entire list, and store its iterator for future use. Before entering the method you check if the availableVms iterator has been prepared. If it has been prepared, grab the next item off of it if it's available; otherwise, return zero.
If the list has not been prepared yet, run your algorithm, and add the results to a temporary list tmp. Once the list is ready, grab its iterator, and use it for subsequent invocations.

How to leftshift an ArrayList

I'm using an ArrayList to hold a history of objects. Each new object I add using the .add method, like:
if(event.getAction() == MotionEvent.ACTION_UP)
{
if(currentWord != null)
{
wordHist.add(currentWord);
}
if(wordHist.size() > WORDHIST_MAX_COUNT)
{
wordHist.remove(0);
}
}
However I don't want this to grow indefinitely, but to be limited to a certain value. If it reaches this maximum value, I want the oldest object (index 0) to be removed, and the rest to be left shifted, so previous index 1 is now index 0, etc.
How can this be done?
Thanks
ArrayList is not really a good choice in this case, but it can by done by calling remove(0) method. But if you want to do that efficiently, a linked list is better
(edited to make it clear that LinkedList is not generally better than ArrayList, but only in this case)
If it reaches this maximum value, I want the oldest object (index 0) to be removed
Then do wordHist.remove(0). That will remove the element at index 0.
To be precise:
wordHist.add(new Word("hello"));
if (wordHist.size() > MAX_SIZE)
wordHist.remove(0);
As user658991 states however, you should be aware of that this is a linear operation, i.e., takes time proportional to the number of elements in the list.
You could do this in constant time using LinkedList methods add and removeFirst.
Another option would be to wrap an array, or ArrayList in a class called something like CircularArrayList. In circular list structures you'll override the oldest element when adding a new one.
Edit:
Your code works fine:
import java.util.*;
class Test {
static int WORDHIST_MAX_COUNT = 3;
static List<String> wordHist = new ArrayList<String>();
public static void add(String currentWord) {
// VERBATIM COPY OF YOUR CODE
if (true/*event.getAction() == MotionEvent.ACTION_UP*/)
{
if(currentWord != null)
{
wordHist.add(currentWord);
}
if(wordHist.size() > WORDHIST_MAX_COUNT)
{
wordHist.remove(0);
}
}
}
public static void main(String[] args) {
add("a");
add("b");
add("c");
for (int i = 0; i < wordHist.size(); i++)
System.out.printf("i: %d, word: %s%n", i, wordHist.get(i));
System.out.println();
add("d");
for (int i = 0; i < wordHist.size(); i++)
System.out.printf("i: %d, word: %s%n", i, wordHist.get(i));
}
}
Prints:
i: 0, word: a
i: 1, word: b
i: 2, word: c
i: 0, word: b <-- b is now at index 0.
i: 1, word: c
i: 2, word: d
Use the remove( ) method.
Using remove(0) will remove the element from the 0th index.
U can use list.remove(index)// here index being '0', this internally shifts rest of the array up. An alternative solution wud be to use a queue or dequeue.
One simple implementation of what Op De Cirkel suggested
import java.util.ArrayList;
import java.util.List;
public class SimpleCircularHistory {
private int sizeLimit, start = 0, end = 0;
boolean empty = false;
private List<String> history;
public SimpleCircularHistory(int sizeLimit) {
this.sizeLimit = sizeLimit;
history = new ArrayList<String>(sizeLimit);
}
public void add(String state){
empty = false;
end = (end + 1) % sizeLimit;
if(history.size() < sizeLimit){
history.add(state);
}else {
history.set(end, state);
start = (end + 1) % sizeLimit;
}
}
public String rollBack(){
if(empty){ // Empty
return null;
}else {
String state = history.get(end);
if(start == end){
empty = true;
}else {
end = (end + sizeLimit - 1) % sizeLimit;
}
return state;
}
}
public void print(){
if(empty){
System.out.println("Empty");
}else {
for(int i = start;; i = (i + 1) % sizeLimit){
System.out.println(history.get(i));
if(i == end) break;
}
System.out.println();
}
}
public static void main(String[] args) {
SimpleCircularHistory h = new SimpleCircularHistory(3);
h.add("a");
h.add("b");
h.add("c");
h.add("d");
h.add("e");
h.add("f");
h.print();
h.add("X");
h.add("Y");
h.rollBack();
h.rollBack();
h.print();
h.add("t");
h.add("v");
h.add("w");
h.print();
h.rollBack();
h.rollBack();
h.rollBack();
h.print();
h.rollBack();
h.print();
}
}
This would print out :
d
e
f
f
t
v
w
Empty
Empty
Yeah, I've noticed this behaviour in adroid's lists too. It's REALLY irritating.
Anyway, there is a way to get around it if I don't mind object creation/destruction and the resulting garbage collection (NEVER do this in a onDraw of a surfaceview or something).
What I do is basically have two tracking int's; one to place the new object, and one to remove it:
int trackInt = 0;
int removeInt = 0;
//and then, in the method/class you use this:
Object newobject = new Object();
//add to list
objectList.add(trackInt, newobject);
trackInt++;
if (bugList.size() > 20) //20 is the max number of object you want, ie the maximum size of the list
{
objectList.remove(removeInt);
trackInt = removeInt;
removeInt++;
if (removeInt > 19) //remember, the list is zero indexed!
{
removeInt = 0;
}
}
Commons-collections has exactly what you're looking for:
http://commons.apache.org/collections/apidocs/org/apache/commons/collections/buffer/CircularFifoBuffer.html

removeAll seems to affect its argument

I have written a generic Partition class (a partition is a division of a set into disjoint subsets, called parts). Internally this is a Map<T,Integer> and a Map<Integer,Set<T>>, where the Integers are the labels of the parts. For example partition.getLabel(T t) gives the label of the part that t is in, and partition.move(T t, Integer label) moves t to the partition labelled by label (internally, it updates both the Maps).
But my method for moving a Collection of objects to a new part does not work. It seems that Set.removeAll() is affecting its argument. I think the problem is something like a ConcurrentModificationException, but I can't work it out. Sorry the code is rather long, but I have marked where the problem is (about half-way down), and the output at the bottom should make it clear what the problem is - at the end the partition is in an illegal state.
import java.util.*;
public class Partition<T> {
private Map<T,Integer> objToLabel = new HashMap<T,Integer>();
private Map<Integer,Set<T>> labelToObjs =
new HashMap<Integer,Set<T>>();
private List<Integer> unusedLabels;
private int size; // = number of elements
public Partition(Collection<T> objects) {
size = objects.size();
unusedLabels = new ArrayList<Integer>();
for (int i = 1; i < size; i++)
unusedLabels.add(i);
// Put all the objects in part 0.
Set<T> part = new HashSet<T>(objects);
for (T t : objects)
objToLabel.put(t,0);
labelToObjs.put(0,part);
}
public Integer getLabel(T t) {
return objToLabel.get(t);
}
public Set<T> getPart(Integer label) {
return labelToObjs.get(label);
}
public Set<T> getPart(T t) {
return getPart(getLabel(t));
}
public Integer newPart(T t) {
// Move t to a new part.
Integer newLabel = unusedLabels.remove(0);
labelToObjs.put(newLabel,new HashSet<T>());
move(t, newLabel);
return newLabel;
}
public Integer newPart(Collection<T> things) {
// Move things to a new part. (This assumes that
// they are all in the same part to start with.)
Integer newLabel = unusedLabels.remove(0);
labelToObjs.put(newLabel,new HashSet<T>());
moveAll(things, newLabel);
return newLabel;
}
public void move(T t, Integer label) {
// Move t to the part "label".
Integer oldLabel = getLabel(t);
getPart(oldLabel).remove(t);
if (getPart(oldLabel).isEmpty()) // if the old part is
labelToObjs.remove(oldLabel); // empty, remove it
getPart(label).add(t);
objToLabel.put(t,label);
}
public void moveAll(Collection<T> things, Integer label) {
// Move all the things from their current part to label.
// (This assumes all the things are in the same part.)
if (things.size()==0) return;
T arbitraryThing = new ArrayList<T>(things).get(0);
Set<T> oldPart = getPart(arbitraryThing);
// THIS IS WHERE IT SEEMS TO GO WRONG //////////////////////////
System.out.println(" oldPart = " + oldPart);
System.out.println(" things = " + things);
System.out.println("Now doing oldPart.removeAll(things) ...");
oldPart.removeAll(things);
System.out.println(" oldPart = " + oldPart);
System.out.println(" things = " + things);
if (oldPart.isEmpty())
labelToObjs.remove(objToLabel.get(arbitraryThing));
for (T t : things)
objToLabel.put(t,label);
getPart(label).addAll(things);
}
public String toString() {
StringBuilder result = new StringBuilder();
result.append("\nPARTITION OF " + size + " ELEMENTS INTO " +
labelToObjs.size() + " PART");
result.append((labelToObjs.size()==1 ? "" : "S") + "\n");
for (Map.Entry<Integer,Set<T>> mapEntry :
labelToObjs.entrySet()) {
result.append("PART " + mapEntry.getKey() + ": ");
result.append(mapEntry.getValue() + "\n");
}
return result.toString();
}
public static void main(String[] args) {
List<String> strings =
Arrays.asList("zero one two three".split(" "));
Partition<String> p = new Partition<String>(strings);
p.newPart(strings.get(3)); // move "three" to a new part
System.out.println(p);
System.out.println("Now moving all of three's part to the " +
"same part as zero.\n");
Collection<String> oldPart = p.getPart(strings.get(3));
//oldPart = Arrays.asList(new String[]{"three"}); // works fine!
p.moveAll(oldPart, p.getLabel(strings.get(0)));
System.out.println(p);
}
}
/* OUTPUT
PARTITION OF 4 ELEMENTS INTO 2 PARTS
PART 0: [two, one, zero]
PART 1: [three]
Now moving all of three's part to the same part as zero.
oldPart = [three]
things = [three]
Now doing oldPart.removeAll(things) ...
oldPart = []
things = []
PARTITION OF 4 ELEMENTS INTO 1 PART
PART 0: [two, one, zero]
*/
Using my debugger I place a breakpoint before the removeAll and I can see (as I suspected) that oldPart and things as the same collection so removing all elements clears that collection.
Your code is extremely confusing but as far as I can work out, oldPart and things are actually the same object. Set.removeAll() certainly doesn't affect its argument unless it is the same object as it's invoked on:
public boolean removeAll(Collection<?> c) {
boolean modified = false;
if (size() > c.size()) {
for (Iterator<?> i = c.iterator(); i.hasNext(); )
modified |= remove(i.next());
} else {
for (Iterator<?> i = iterator(); i.hasNext(); ) {
if (c.contains(i.next())) {
i.remove();
modified = true;
}
}
}
return modified;
}
Update:
The easy way to eliminate this is to use a copy of things in the moveAll() method. Indeed such a copy already exists.
T arbitraryThing = new ArrayList<T>(things).get(0);
This line creates but then instantly discards a copy of things. So I'd suggest replacing it with:
ArrayList<T> thingsToRemove = new ArrayList<T>(things)
T arbitraryThing = thingsToRemove.get(0);
And in the rest of the method, replace all references to things to thingsToRemove.

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