java.util.ConcurrentModificationException 异常解决办法及原理

最近在修程序的bug，发现后台抛出以下异常：

Exception in thread "main" java.util.ConcurrentModificationException
at java.util.HashMap$HashIterator.nextEntry(HashMap.java:793)
at java.util.HashMap$KeyIterator.next(HashMap.java:828)
at com.keyman.demo.test.ClearResultTable.method2(ClearResultTable.java:54)
at com.keyman.demo.test.ClearResultTable.main(ClearResultTable.java:88)

找到报错行：at com.keyman.demo.test.ClearResultTable.method2(ClearResultTable.java:54)发现，报错位置：for (String s1 : sets)

Set<String> sets = map.keySet();
for (String s1 : sets) {
String value = map.get(s1);
// 删除满足value以abc开头的键值对
if (value.startsWith("abc")) {
map.remove(s1);
}
}

或者下面的方式同样也会抛出异常

Iterator<String> iterator = map.keySet().iterator();
while (iterator.hasNext()) {
String key = iterator.next();
String value = map.get(key);
// 删除满足value以abc开头的键值对
if (value.startsWith("abc")) {
map.remove(key);
//iterator.remove(); // 同步modCount和expectedModCount

}
}

其实不管是Map还是Set这样操作时均会抛出此异常！

解决办法为：如果不是Iterator迭代方式，则修改map迭代方式为Iterator()方式，采用iterator.remove();而不直接通过map.remove();

Iterator<String> iterator = map.keySet().iterator();
while (iterator.hasNext()) {
String key = iterator.next();
String value = map.get(key);
// 删除满足value以abc开头的键值对
if (value.startsWith("abc")) {
//map.remove(value);
iterator.remove();   // 关键代码，同步modCount和expectedModCount

}
}

详细原因如下：

发现这个位置应该是不会报错的，查找前后文，发现最有可能报错的应该是for循环里面，但是咋一看压根没错！通过查找资料发现：当修改的个数跟期望修改的个数不相等时抛出此异常。

private abstract class HashIterator<E> implements Iterator<E> {
Entry<K, V> next; // next entry to return
int expectedModCount; // For fast-fail
int index; // current slot
Entry<K, V> current; // current entry
...
final Entry<K, V> nextEntry() {
if (modCount != expectedModCount)
throw new ConcurrentModificationException(); // 抛出异常
Entry<K, V> e = current = next;
if (e == null)
throw new NoSuchElementException();

if ((next = e.next) == null) {
Entry[] t = table;
while (index < t.length && (next = t[index++]) == null)
;
}
return e;
}
...
}

于是查看HashMap.remove()方法代码如下：

/**
* Removes the mapping for the specified key from this map if present.
*
* @param  key key whose mapping is to be removed from the map
* @return the previous value associated with <tt>key</tt>, or
*         <tt>null</tt> if there was no mapping for <tt>key</tt>.
*         (A <tt>null</tt> return can also indicate that the map
*         previously associated <tt>null</tt> with <tt>key</tt>.)
*/
public V remove(Object key) {
Entry<K,V> e = removeEntryForKey(key);
return (e == null ? null : e.value);
}

/**
* Removes and returns the entry associated with the specified key
* in the HashMap.  Returns null if the HashMap contains no mapping
* for this key.
*/
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++;
size--;
if (prev == e)
table[i] = next;
else
prev.next = next;
e.recordRemoval(this);
return e;
}
prev = e;
e = next;
}

return e;
}

你会发现，其中有modCount++操作。
modCount表示修改的次数，而并没有改变其exceptedmodCount；

接下来看看iterator.remove()方法：（java.util.Hashtable.Enumerator.remove()）

public void remove() {
if (!iterator)
throw new UnsupportedOperationException();
if (lastReturned == null)
throw new IllegalStateException("Hashtable Enumerator");
if (modCount != expectedModCount)
throw new ConcurrentModificationException();

synchronized(Hashtable.this) {
Entry[] tab = Hashtable.this.table;
int index = (lastReturned.hash & 0x7FFFFFFF) % tab.length;

for (Entry<K,V> e = tab[index], prev = null; e != null;
prev = e, e = e.next) {
if (e == lastReturned) {
modCount++;
expectedModCount++;
if (prev == null)
tab[index] = e.next;
else
prev.next = e.next;
count--;
lastReturned = null;
return;
}
}
throw new ConcurrentModificationException();
}
}
}

而此删除元素的方法，将modCount自增的同时将exceptedModCount同样自增。也就不会抛出异常。

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