java遍历集合删除特定元素的 java.util.ConcurrentModificationException异常

看一段代码：

for (SecurityPsn psn : psnList) {
   if(StringUtils.isBlank(psn.getCertCode())){
   psnList.remove(psn);
  }
 }

该代码会抛出java.util.ConcurrentModificationException异常

 

正确代码为：

Iterator<SecurityPsn> iter = psnList.iterator(); 
         while (iter.hasNext()) { 
               SecurityPsn psn = iter.next(); 
              if (StringUtils.isBlank(psn.getCertCode())) { 
                   iter.remove(); 
              } 
           } 

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

原因：当修改的个数跟期望修改的个数不相等时抛出此异常。

如果改成：

for (int i=0;i<psnList.size();i++ ) {
   if(StringUtils.isBlank(psnList.get(i).getCertCode())){
   psnList.remove(psn);
  }
 }

虽然不报上述错误，但结果不准确，且最后一个下标可能会越界。

因为集合的大小是动态变化的，当你删除一个元素之后，元素中的序号又重新排列，原来第二个应该删除的元素现在排在了第一个元素的位置，真正删除的却是第三个元素，依次类推，删除的是第一个、第三个、第五个,所以第二种正确的做法为：

for (int i=0;i<psnList.size();i++ ) {
   if(StringUtils.isBlank(psnList.get(i).getCertCode())){
   psnList.remove(psnList.get(i));

 i--;
  }
 }

源码分析：

  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;
  }
  ...

查看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;
  }

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。