Java Concurrentmodificationexception Exception Analysis and Solutions

Java Concurrentmodificationexception Exception Analysis and Solutions
Http://www.2cto.com/kf/201403/286536.html

Java.util.ConcurrentModificationException Solutions
http://blog.csdn.net/lipei1220/article/details/9028669

Cause: Iterator do traversal, HashMap is modified (Bb.remove (ele), size-1), Iterator (Object ele=it.next ()) will check HashMap size,size change, Throws an error concurrentmodificationexception.

// 1 使用Iterator提供的remove方法，用于删除当前元素   for (Iterator<string> it = myList.iterator(); it.hasNext();) {       String value = it.next();        if (value.equals( "3" )) {            it.remove();  // ok       } } System. out.println( "List Value:" + myList.toString());  // 2 建一个集合，记录需要删除的元素，之后统一删除             List<string> templist = new ArrayList<string>();   for (String value : myList) {        if (value.equals( "3" )) {            templist.remove(value);       } }   // 可以查看removeAll源码，其中使用Iterator进行遍历 myList.removeAll(templist); System. out.println( "List Value:" + myList.toString());          // 3. 使用线程安全CopyOnWriteArrayList进行删除操作 List<string> myList = new CopyOnWriteArrayList<string>(); myList.add( "1" ); myList.add( "2" ); myList.add( "3" ); myList.add( "4" ); myList.add( "5" ); Iterator<string> it = myList.iterator();   while (it.hasNext()) {       String value = it.next();        if (value.equals( "3" )) {            myList.remove( "4" );            myList.add( "6" );            myList.add( "7" );       } } System. out.println( "List Value:" + myList.toString());  // 4. 不使用Iterator进行遍历，需要注意的是自己保证索引正常   for ( int i = 0 ; i < myList.size(); i++) {       String value = myList.get(i);       System. out.println( "List Value:" + value);        if (value.equals( "3" )) {            myList.remove(value);  // ok            i--; // 因为位置发生改变，所以必须修改i的位置       } } System. out.println( "List Value:" + myList.toString());

Java Concurrentmodificationexception Exception Analysis and Solutions