迭代中增删Java集合中对象的正确方法
看下面这个例子:
@Test public void listRemoveTest() { String[] temp={"Jim","Jim","Amli","Amli","Masu","Lina"}; List<String> names = new ArrayList<String>(); for (int i = 0; i < temp.length; i++) { names.add(temp[i]); } for (int i = 0; i < names.size(); i++) { String name = names.get(i); if("Jim".equals(name) || "Amli".equals(name)) { names.remove(i); } } for (String name : names) { System.out.println(name); } }
我们的期望值是移除所有名为“Jim”或者“Amli”的对象:
Masu Lina
而事实上输出结果为:
Jim Amli Masu Lina
造成该结果是由于 移除列表中对象后 列表长多缩短而列表索引值 i 没有做相应的调整,可以将代码修改如下:
for (int i = 0; i < names.size();) { String name = names.get(i); if("Jim".equals(name) || "Amli".equals(name)) { names.remove(i); continue; } else { i++; } }
还可以用集合的迭代器来处理,代码如下:
//错误处理方式 for (String name : names) { if("Jim".equals(name) || "Amli".equals(name)) { names.remove(name); } } //正确处理方式 Iterator<String> iterator = names.iterator(); while (iterator.hasNext()) { String name = iterator.next(); if("Jim".equals(name) || "Amli".equals(name)) { iterator.remove(); } }
上述的错误方式会造成java.util.ConcurrentModificationException。
同样的问题也存在于Map中。例如下面的代码:
@Test public void removeTest() { Map<String, String> params = new HashMap<String, String>(); params.put("STUDENT_A", "S.A"); params.put("STUDENT_B", "S.B"); params.put("TEACHER_A", "T.A"); params.put("TEACHER_B", "T.B"); params.put("TEST", "TEST"); try { removeTeacherParameters(params); } catch (Exception e) { e.printStackTrace(); } } //错误的使用方式 private void removeTeacherParameters(Map<String, String> params) { for(String key:params.keySet()){ if(key.equals("TEST")){ params.remove(key); } if(key.indexOf("TEACHER_")==-1){ params.remove(key); } } }
这里会出现java.util.ConcurrentModificationException:
java.util.ConcurrentModificationException at java.util.HashMap$HashIterator.nextEntry(HashMap.java) at java.util.HashMap$KeyIterator.next(HashMap.java)
出现这个问题原因是由于 集合中的修改次数标量 和 集合的迭代器的期望修改次数标量 没有匹配上造成的具体可以看HashMap中的代码:
public V remove(Object key) { Entry<K,V> e = removeEntryForKey(key); return (e == null ? null : e.value); } final Entry<K,V> removeEntryForKey(Object key) { int hash = (key == null) ? 0 : hash(key.hashCode()); int i = indexFor(hash, table.length); Entry<K,V> prev = table[i]; Entry<K,V> e = prev; while (e != null) { Entry<K,V> next = e.next; Object k; if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) { modCount++; //只增加了modCount size--; if (prev == e) table[i] = next; else prev.next = next; e.recordRemoval(this); return e; } prev = e; e = next; } return e; }
而AbstractHashedMap中的代码显示在进行next()操作的时候会检测modCount 是否和 expectedModCount相等:
public class AbstractHashedMap extends AbstractMap implements IterableMap { protected transient int modCount; protected static abstract class HashIterator implements Iterator { protected int expectedModCount; public boolean hasNext() { return (next != null); } protected HashEntry nextEntry() { //当expectedModCount和modCount不相等时,就抛出ConcurrentModificationException<br> if (parent.modCount != expectedModCount) { throw new ConcurrentModificationException(); } HashEntry newCurrent = next; if (newCurrent == null) { throw new NoSuchElementException(AbstractHashedMap.NO_NEXT_ENTRY); } HashEntry[] data = parent.data; int i = hashIndex; HashEntry n = newCurrent.next; while (n == null && i > 0) { n = data[--i]; } next = n; hashIndex = i; last = newCurrent; return newCurrent; } } protected static class HashMapIterator extends HashIterator implements MapIterator { public Object next() { return super.nextEntry().getKey(); } } }
而正确的使用方法如下:
@Test public void removeTest() { Map<String, String> params = new HashMap<String, String>(); params.put("STUDENT_A", "S.A"); params.put("STUDENT_B", "S.B"); params.put("TEACHER_A", "T.A"); params.put("TEACHER_B", "T.B"); params.put("TEST", "TEST"); try { removeStudentParameters(params); } catch (Exception e) { e.printStackTrace(); } } //正确的使用方式 private void removeStudentParameters(Map<String, String> params) { Iterator<Entry<String, String>> it = params.entrySet().iterator(); while (it.hasNext()) { String key = it.next().getKey(); if(key.equals("TEST") || key.indexOf("STUDENT_")==-1){ it.remove(); } } }
这是由于在迭代器在进行remove操作时候, 同步了modCount 和 expectedModCount的值:
public class AbstractHashedMap extends AbstractMap implements IterableMap { protected transient int modCount; protected static abstract class HashIterator implements Iterator { protected int expectedModCount; public void remove() { if (last == null) { throw new IllegalStateException(AbstractHashedMap.REMOVE_INVALID); } if (parent.modCount != expectedModCount) { throw new ConcurrentModificationException(); } parent.remove(last.getKey()); last = null; //重新设置了expectedModCount的值,避免了ConcurrentModificationException的产生 expectedModCount = parent.modCount; } } }
产生ConcurrentModificationException的原因就是:
执行remove(Object o)方法之后,modCount和expectedModCount不相等了。然后当代码执行到next()方法时,判断了checkForComodification(),发现两个数值不等,就抛出了该Exception。要避免这个Exception,就应该使用结合内置迭代器的remove()方法。同理,在进行集合迭代的时候往集合中插入新的元素也会造成同样的问题,要解决这个问题是新建一个集合来处理。
参考:
http://*.com/questions/602636/concurrentmodificationexception-and-a-hashmap
http://www.blogjava.net/evanliu/archive/2008/08/31/224453.html
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