这个为什么要单独说的原因是,在开发中的对数据库中的增删为最基本的,但是是不是写对了就尤为重要
先来看代码:
1 public void testLoopInList(){ 2 List<String> a = new ArrayList<String>(); 3 a.add("1"); 4 a.add("2"); 5 a.add("w"); 6 for (String string : a) { 7 System.out.println(string); 8 } 9 10 for (String temp : a) { 11 if("2".equals(temp)){ 12 a.remove(temp); 13 } 14 } 15 16 for (String string : a) { 17 System.out.println(string); 18 }
输出:
1 2 java.util.ConcurrentModificationException at java.util.ArrayList$Itr.checkForComodification(ArrayList.java:859) at java.util.ArrayList$Itr.next(ArrayList.java:831) at test.Test_ForeachAndIterator.testLoopInList(Test_ForeachAndIterator.java:19)
出现错误了
原因:首先从错误中可以看出,是ArraList中的Iterator的checkForComodification()出现的错误,说明了foreach的实现原理其实就是实现了内部类I特Iterator来进行遍历的,其次为什么会出错呢?
因为在ArrayList和ArrayList的内部类的Iterator中的都会存在remove的方法,而ArrayList和Iterator中都各自有自己的维持长度的变量,前者是modCount,后者是expectModCount,但是源码中ArrayList的remove方法是会改变modCount的值,但却不会直接同步到expectModCount的值的,而Iterator中时刻都插入了checkForComodification()方法来监测modCount是否与expectModCount相等,所以就会很容易出现异常错误,当然下面的代码也是错的
1 public void testLoopInList(){ 2 List<String> a = new ArrayList<String>(); 3 a.add("1"); 4 a.add("2"); 5 6 for (String string : a) { 7 System.out.println(string); 8 } 9 10 Iterator<String> it = a.iterator(); 11 while(it.hasNext()){ 12 if("1".equals(it.next())){ 13 a.remove(it.next()); 14 } 15 } 16 17 for (String string : a) { 18 System.out.println(string); 19 } 20 }
错误原因同上
所以在解决问题的关键就是要避免这个异常的出现,也就是时刻让modCount==expectModCount,所以就是使用iterator的remove方法,因为会有使两者相等的代码
即下面:
1 public void testLoopInList(){ 2 List<String> a = new ArrayList<String>(); 3 a.add("1"); 4 a.add("2"); 5 6 for (String string : a) { 7 System.out.println(string); 8 } 9 10 Iterator<String> it = a.iterator(); 11 while(it.hasNext()){ 12 if("1".equals(it.next())){ 13 it.remove(); 14 } 15 } 16 17 for (String string : a) { 18 System.out.println(string); 19 } 20 }
相关源码提上:
1 private class Itr implements Iterator<E> { 2 /** 3 * Index of element to be returned by subsequent call to next. 4 */ 5 int cursor = 0; 6 /** 7 * Index of element returned by most recent call to next or 8 * previous. Reset to -1 if this element is deleted by a call 9 * to remove. 10 */ 11 int lastRet = -1; 12 /** 13 * The modCount value that the iterator believes that the backing 14 * List should have. If this expectation is violated, the iterator 15 * has detected concurrent modification. 16 */ 17 int expectedModCount = modCount; 18 public boolean hasNext() { 19 return cursor != size(); 20 } 21 public E next() { 22 checkForComodification(); 23 try { 24 E next = get(cursor); 25 lastRet = cursor++; 26 return next; 27 } catch (IndexOutOfBoundsException e) { 28 checkForComodification(); 29 throw new NoSuchElementException(); 30 } 31 } 32 public void remove() { 33 if (lastRet == -1) 34 throw new IllegalStateException(); 35 checkForComodification(); 36 try { 37 AbstractList.this.remove(lastRet); 38 if (lastRet < cursor) 39 cursor--; 40 lastRet = -1; 41 expectedModCount = modCount; 42 } catch (IndexOutOfBoundsException e) { 43 throw new ConcurrentModificationException(); 44 } 45 } 46 final void checkForComodification() { 47 if (modCount != expectedModCount) 48 throw new ConcurrentModificationException(); 49 }
ArrayList中的remove
1 public boolean remove(Object o) { 2 if (o == null) { 3 for (int index = 0; index < size; index++) 4 if (elementData[index] == null) { 5 fastRemove(index); 6 return true; 7 } 8 } else { 9 for (int index = 0; index < size; index++) 10 if (o.equals(elementData[index])) { 11 fastRemove(index); 12 return true; 13 } 14 } 15 return false; 16 }
当然你说我不用for增强,就用普通的for ,就不会有创建Iterator的操作,进而就不会有Iterator的实时监测维护值得操作,也行啊,但还是要注意有坑啊! 比如我们来看下面的代码:
1 List list = new ArrayList(); 2 Collections.addAll(list, 1, 2, 3, 4, 5); 3 for(int i = 0; i < list.size(); ++i){ 4 int val = (int) list.get(i); 5 6 // 需求是删除 3 和 4 两个元素 7 if(3 == val || 4 == val){ 8 list.remove(i); 9 } 10 }
这是我们非常常见的写法,结果 为 :1 2 3 4
奇怪了,4为甚没有删除掉,其实仔细想想 还是能相同的 删除了3之后 i = 3 size = 4,这时候,list.get返回值为5, 会跳过4
所以 修改的话,两个 liter tips : 要么自己维护索引,要么反向循环
1 for(int i = list.size() - 1; i > 0; --i){ 2 int val = (int) list.get(i); 3 4 // 需求是删除 3 和 4 两个元素 5 if(3 == val || 4 == val){ 6 list.remove(i); 7 } 8 }
1 for(int i = 0; i < list.size(); ++i){ 2 int val = (int) list.get(i); 3 4 // 需求是删除 3 和 4 两个元素 5 if(3 == val || 4 == val){ 6 list.remove(i); 7 i --; // 删除后,索引也应该变化 8 } 9 }
ok, that's all ...