blocks|key|527358|text|只要你不要求"remove“方法是原子的，这应该没问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|527359|换句话说，如果“做某事”检查项目在列表中出现不止一次，那么当您到达下一行时，该检查的结果可能是错误的。|527360|此外，确保在迭代时在列表上进行同步：|527361|synchronized(list)+{
++++for+(Object+o+:+list)+{}
}|code-block|syntax|javascript|527362|正如Peter+Lawrey所提到的，CopyOnWriteArrayList可以让您的工作更轻松，并且可以在高度并发的环境中提供更好的性能。|offset|length|527363|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/util/concurrent/CopyOnWriteArrayList.html^0|0|0|0|0|J|K|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

That should be fine as long as you don't require the "remove" method to be atomic.

In other words, if the "do something" checks that the item appears more than once in the list for example, it is possible that the result of that check will be wrong by the time you reach the next line.

Also, make sure you synchronize on the list when iterating:

<pre><code>synchronized(list) {
 for (Object o : list) {}
}
</code></pre>

As mentioned by Peter Lawrey, <a href="http://docs.oracle.com/javase/7/docs/api/java/util/concurrent/CopyOnWriteArrayList.html">CopyOnWriteArrayList</a> can make your life easier and can provide better performance in a highly concurrent environment.

blocks|key|527414|text|它将为添加/删除操作提供一致的行为。但在迭代时，您必须显式同步。Refer+this+link|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|527415|entityMap|0|LINK|mutability|MUTABLE|url|http://bmraoblog.blogspot.in/2006/11/using-collectionssynchronizedlist.html^0|W|F|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

It will give consistent behavior for add/remove operations. But while iterating you have to explicitly synchronized. <a href="http://bmraoblog.blogspot.in/2006/11/using-collectionssynchronizedlist.html" rel="nofollow">Refer this link</a>

blocks|key|310476|text|是的，它会工作的很好，因为你有synchronized的列表。我建议你使用CopyOnWriteArrayList。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|310477|CopyOnWriteArrayList<String>+cpList=new+CopyOnWriteArrayList<String>(new+ArrayList<String>());

++++void+remove(String+item)
++++{
+++++++++do+something;+(doesn't+work+on+the+list)
+++++++++++++++++cpList..remove(item);
++++}|code-block|syntax|javascript|310478|entityMap^0|F|C|11|K|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Yes, it will work fine as you have <code>synchronized</code> the list . I would suggest you to use <code>CopyOnWriteArrayList</code>.

<pre><code>CopyOnWriteArrayList&lt;String&gt; cpList=new CopyOnWriteArrayList&lt;String&gt;(new ArrayList&lt;String&gt;());

 void remove(String item)
 {
 do something; (doesn't work on the list)
 cpList..remove(item);
 }
</code></pre>

blocks|key|310508|text|列表可能有两个不同的问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|310509|1)如果您在迭代中进行修改，即使是在单线程环境中，您也将拥有如下示例中的ConcurrentModificationException：|310510|List<String>+list+=+new+ArrayList<String>();
for+(int+i=0;i<5;i%2B%2B)
+++list.add("Hello+"%2Bi);

for(String+msg:list)
+++list.remove(msg);|code-block|syntax|javascript|310511|因此，为了避免这个问题，您可以这样做：|310512|for(int+i=list.size()-1;i>=0;i--)
+++list.remove(i);|310513|2)第二个问题可能是多线程环境。如上所述，您可以使用同步(列表)来避免异常。|310514|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You can have 2 diffent problems with lists : 
1) If you do a modification within an iteration even though in a mono thread environment, you will have ConcurrentModificationException like in this following example :

<pre><code>List&lt;String&gt; list = new ArrayList&lt;String&gt;();
for (int i=0;i&lt;5;i++)
 list.add("Hello "+i);

for(String msg:list)
 list.remove(msg);
</code></pre>

So, to avoid this problem, you can do :

<pre><code>for(int i=list.size()-1;i&gt;=0;i--)
 list.remove(i);
</code></pre>

2)The second problem could be multi threading environment. As mentioned above, you can use synchronized(list) to avoid exceptions.

blocks|key|310521|text|synchronized(list)+{
++++for+(Object+o+:+list)+{}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|310522|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>synchronized(list) {
 for (Object o : list) {}
}
</code></pre>

I have a pre-populated array list. And I have multiple threads which will remove elements from the array list. Each thread calls the remove method below and removes one item from the list. Does the following code give me consistent behavior ?

<pre><code>ArrayList&lt;String&gt; list = Collections.synchronizedList(new ArrayList&lt;String&gt;());

void remove(String item)
{
 do something; (doesn't work on the list)
 list.remove(item);
}
</code></pre>

Thanks!

Java Synchronized list

我有一个预先填充的数组列表。并且我有多个线程可以从数组列表中删除元素。每个线程调用下面的remove方法，并从列表中删除一项。下面的代码给了我一致的行为吗？ArrayList<String> list = Collections.synchronizedList(new ArrayList<String>());voi...

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