Java并发(五):并发,迭代器和容器

　　在随后的博文中我会继续分析并发包源码,在这里,得分别谈谈容器类和迭代器及其源码,虽然很突兀,但我认为这对于学习Java并发很重要;

ConcurrentModificationException:

　　JavaAPI中的解释:当不允许这样的修改时，可以通过检测到对象的并发修改的方法来抛出此异常。一个线程通常不允许修改集合，而另一个线程正在遍历它。 一般来说，在这种情况下，迭代的结果是未定义的。 某些迭代器实现（包括由JRE提供的所有通用集合实现的实现）可能会选择在检测到此行为时抛出此异常。 这样做的迭代器被称为"及时失败"迭代器，当他们发现容器在迭代时被修改时,就会报异常;它称不上时一种处理机制,而是一种预防机制,只能作为并发问题的预警指示器.

迭代器与容器:

　　Vector这个"古老"的容器类相信大家很熟悉了,他是线程安全的没错,我们利用elements()方法遍历,不会出现线程安全的问题:

 /**
   *  JDK1.8源码
 */

 /**
      * Returns an enumeration of the components of this vector. The
      * returned {@code Enumeration} object will generate all items in
      * this vector. The first item generated is the item at index {@code 0},
      * then the item at index {@code 1}, and so on.
      *
      * @return  an enumeration of the components of this vector
      * @see     Iterator
      */
     public Enumeration<E> elements() {
         return new Enumeration<E>() {
             int count = 0;

             public boolean hasMoreElements() {
                 return count < elementCount;
             }

             public E nextElement() {
                 synchronized (Vector.this) {
                     if (count < elementCount) {
                         return elementData(count++);
                     }
                 }
                 throw new NoSuchElementException("Vector Enumeration");
             }
         };
     }

JDK1.8源码: AbstractCollection (been extends by Vector)

　　但当我们利用foreach进行迭代时,底层自动调用了iterator(),若我们不锁住容器,可能会报ConcurrentModificationException

 /**
      * Returns an iterator over the elements in this list in proper sequence.
      *
      * <p>The returned iterator is <a href="#fail-fast"><i>fail-fast</i></a>.
      *
      * @return an iterator over the elements in this list in proper sequence
      */
     public Iterator<E> iterator() {
         return new Itr();
         //JDK1.8写成了一个内部类的形式返回迭代器
     }

     /**
      * An optimized version of AbstractList.Itr
      */
     private class Itr implements Iterator<E> {
         int cursor;       // index of next element to return
         int lastRet = -1; // index of last element returned; -1 if no such
         int expectedModCount = modCount;

         Itr() {}

         public boolean hasNext() {
             return cursor != size;
         }

         //观察下面的代码可以发现,每次迭代一次都要检查容器长度是否改变
         @SuppressWarnings("unchecked")
         public E next() {
             checkForComodification();
             int i = cursor;
             if (i >= size)
                 throw new NoSuchElementException();
             Object[] elementData = ArrayList.this.elementData;
             if (i >= elementData.length)
                 throw new ConcurrentModificationException();
             cursor = i + 1;
             return (E) elementData[lastRet = i];
         }

         public void remove() {
             if (lastRet < 0)
                 throw new IllegalStateException();
             checkForComodification();

             try {
                 ArrayList.this.remove(lastRet);
                 cursor = lastRet;
                 lastRet = -1;
                 expectedModCount = modCount;
             } catch (IndexOutOfBoundsException ex) {
                 throw new ConcurrentModificationException();
             }
         }

         @Override
         @SuppressWarnings("unchecked")
         public void forEachRemaining(Consumer<? super E> consumer) {
             Objects.requireNonNull(consumer);
             final int size = ArrayList.this.size;
             int i = cursor;
             if (i >= size) {
                 return;
             }
             final Object[] elementData = ArrayList.this.elementData;
             if (i >= elementData.length) {
                 throw new ConcurrentModificationException();
             }
             while (i != size && modCount == expectedModCount) {
                 consumer.accept((E) elementData[i++]);
             }
             // update once at end of iteration to reduce heap write traffic
             cursor = i;
             lastRet = i - 1;
             checkForComodification();
         }

         final void checkForComodification() {
             if (modCount != expectedModCount)
                 throw new ConcurrentModificationException();
         }
     }    

JDK1.8源码: Vector

　　查看代码时可以发现,iterator()的内部类中提供的next,remove等方法都进行了迭代操作,这样的迭代被称为"隐藏迭代器";

　　部分容器类的toString()方法会在调用StringBuilder的append()同时迭代容器,例如继承了AbstractSet的HashSet类,而AbstractSet又继承于AbstractCollection:

//  String conversion

    /**
     * Returns a string representation of this collection.  The string
     * representation consists of a list of the collection's elements in the
     * order they are returned by its iterator, enclosed in square brackets
     * (<tt>"[]"</tt>).  Adjacent elements are separated by the characters
     * <tt>", "</tt> (comma and space).  Elements are converted to strings as
     * by {@link String#valueOf(Object)}.
     *
     * @return a string representation of this collection
     */
    public String toString() {
        Iterator<E> it = iterator();
        if (! it.hasNext())
            return "[]";

        StringBuilder sb = new StringBuilder();
        sb.append('[');
        for (;;) {
            E e = it.next();
            sb.append(e == this ? "(this Collection)" : e);
            if (! it.hasNext())
                return sb.append(']').toString();
            sb.append(',').append(' ');
        }
    }

JDK1.8源码: AbstractCollection

　　隐藏的迭代器普遍存在,甚至出现在hashCode,equals,contains,remove等方法中,此处不再赘述;

解决办法:

　　考虑到并发安全性,我们不得不对容器类单独加锁,但同时我们又不希望加锁,那样太损耗性能了,还存在线程饥饿和死锁的风险,极大降低吞吐量和CPU利用率;

　　作为有限的替代,我们可以考虑"克隆"容器,并在将其副本封闭起来进行迭代,避免了抛出ConcurrentModificationException,但在克隆过程中仍然要对容器加锁;显然,克隆容器存在系统开销,我们在选择这种方案时必须考虑的因素有:容器大小,每个元素上执行的工作,迭代操作相对于其它操作的调用频率,以及相应时间和系统吞吐率的需求,具体应用详见CopyOnWriteArrayList类.

参考材料:

　　Java并发编程实战