问题
I user sun jdk 1.5 ThreadPoolExecutor( 24, 24,60,TimeUnit.SECONDS, new LinkedBlockingQueue()). soemtime I use jdb tool to find the status of all threads in thread pool are " waiting in a monitor", the code is :
String key = getKey(dt.getPrefix(), id);
synchronized (key.intern()) { ----->
Is there a problem in "synchronized (key.intern()) " ?
I get following informatnio using jdb tool, the status of 24 threads is "waiting in a monitor", it means 24 threads are deadlock at "key.intern()".
(java.lang.Thread)0x28 pool-3-thread-2 waiting in a monitor
(java.lang.Thread)0x27 pool-3-thread-3 waiting in a monitor
(java.lang.Thread)0x1b pool-3-thread-4 waiting in a monitor
(java.lang.Thread)0x1a pool-3-thread-5 waiting in a monitor
(java.lang.Thread)0x19 pool-3-thread-6 waiting in a monitor
(java.lang.Thread)0x18 pool-3-thread-7 waiting in a monitor
(java.lang.Thread)0x17 pool-3-thread-8 waiting in a monitor ...
so the result is : in multi-threads environment, Sting intern() method may be deadlock, ok ?
回答1:
I posted a related question to this once that you might want to take a look at: Problem with synchronizing on String objects?
What I learned was: using intern'ed Strings for synchronization is a bad practice.
回答2:
Quite. The problem is that key.intern() isn’t really that unique because it’s returning a string from a pool. String.intern() might return the same object even when used on different objects. Try using key itself or a different object altogether.
回答3:
The code is almost certainly trying to synchronize actions that affect the same key. So it's calling intern() to ensure that the same key gets mapped to the same object, and therefore is valid as an object for synchronization.
The problem, if you're running into a bottleneck there (it's not a deadlock) is that you have too many operations coming in at the same time using the same key.
Rethink what needs to be synchronized.
回答4:
If you need to synchronize on a String, don't use a String instance as the mutex (interned or not). A string can be used to create a good mutex object, though: synchronizing on an ID.
回答5:
You are having two problems. The one is using String as the lock. The second one is deadlock.
If you using String as lock, you will lose the control of "who" and "where" will take that object lock.
Your deadlock issue, which may or may not caused by lock on String. However, the actual reason of deadlock is: "Your code can lead deadlock.". If it can happen, it will happen.
You must trace your threads' stacks to resolve deadlocks.
回答6:
There is not enough code here to tell what is going wrong. It could be a bottleneck as has been mentioned, but at least one thread should be running (with fairly heavy CPU usage) for that to happen, or a thread that has the lock is put to sleep without releasing the lock.
A deadlock is another possibility but that would require synchronization on two separate locks on multiple threads, and you have shown only one lock object here.
It is really impossible to determine without more information.
回答7:
As Bombe says, key.intern() won't necessarily give you a very unique key to synchronize on.
You should be cautious about changing the code, however. You need to understand the locking strategy in the code before changing it. Removing the intern() call may give you code that appears to work correctly but contains a data race that will bite you later.
回答8:
You very likely have a deadlock.
If you want to avoid deadlocks, every thread must always acquire locks in the same order. When you use String.intern() to get your locks, you are locking on an instance that any code in the entire JVM has access to, and lock on. Most likely, other threads in your own code are deadlocking, but it doesn't have to be.
I'm not sure what you mean in your answer by "key.intern() guarantee uniqueness". The intern() method reduces uniqueness by returning the same object for every string that's equivalent.
String s1 = new String(new char[] { 'c', 'o', 'm', 'm', 'o', 'n' }).intern();
String s2 = new String("commo" + (s1.charAt(s1.length() - 1)).intern();
String s3 = "common";
if ((s1 == s2) && (s1 == s3))
System.out.println("There's only one object here.");
The code above will demonstrate that even though you created two unique instances, by interning them, you replaced them with a single, canonical instance.
There's danger any time you use an object that's visible outside your own code as a lock. Try to stick to private members, objects that you don't allow to escape from your own stack, etc.
回答9:
How about using a unique string prefix with the locking value and using the String.intern() in the synchronized block. For example if you want to lock on the string "lock1", use a UUID prefix like this: "85e565b3-d440-46e7-93b6-69ee7e9a63ee-lock1". This type of string should not be already in the intern pool. i.e. chance of deadlock by other code is very low.
回答10:
key.intern() guarantee uniqueness because key.intern() returns a string from String constants pool.
http://java.sun.com/j2se/1.4.2/docs/api/java/lang/String.html#intern() intern
public String intern() Returns a canonical representation for the string object. A pool of strings, initially empty, is maintained privately by the class String.
回答11:
String.intern() is a native method - that might be a cause of the problem.
来源：https://stackoverflow.com/questions/348985/deadlock-on-synchronized-string-intern