在本地资源上同步时Java死锁?

Ro.*_*Ro. 3 java concurrency multithreading deadlock thread-safety

我看到同一行代码中多个线程死锁的问题.我不能在本地或任何测试中重现问题,但是生产中的线程转储已经非常清楚地显示了问题.

我无法理解为什么线程会在下面的同步线上被阻塞,因为在调用堆栈或任何其他线程中的Object上没有其他同步.有没有人知道发生了什么,或者我怎么能重现这个问题(目前尝试15个线程都在循环中击中trim(),同时通过我的队列处理2000个任务 - 但无法重现)

在下面的线程转储中,我认为具有"锁定"状态的多个线程可能是Java Bug的一个表现形式:http://bugs.java.com/view_bug.do?bug_id = 8047816其中JStack报告线程处于错误状态.(我使用的是JDK版本:1.7.0_51)

干杯!

这是一个线程转储中的线程视图.....

"xxx>Job Read-3" daemon prio=10 tid=0x00002aca001a6800 nid=0x6a3b waiting for monitor entry [0x0000000052ec4000]
   java.lang.Thread.State: BLOCKED (on object monitor)
    at com.mycompany.collections.CustomQueue.remove(CustomQueue.java:101)
    - locked <0x00002aae6465a650> (a java.util.ArrayDeque)
    at com.mycompany.collections.CustomQueue.trim(CustomQueue.java:318)
    at com.mycompany.collections.CustomQueue.itemProcessed(CustomQueue.java:302)
    at com.mycompany.collections.CustomQueue.trackCompleted(CustomQueue.java:147)
    at java.util.concurrent.ThreadPoolExecutor.afterExecute(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at java.lang.Thread.run(Unknown Source)

   Locked ownable synchronizers:
    - <0x00002aaf5f9c2680> (a java.util.concurrent.ThreadPoolExecutor$Worker)

"xxx>Job Read-2" daemon prio=10 tid=0x00002aca001a5000 nid=0x6a3a waiting for monitor entry [0x0000000052d83000]
   java.lang.Thread.State: BLOCKED (on object monitor)
    at com.mycompany.collections.CustomQueue.remove(CustomQueue.java:101)
    -  locked <0x00002aae6465a650> (a java.util.ArrayDeque)
    at com.mycompany.collections.CustomQueue.trim(CustomQueue.java:318)
    at com.mycompany.collections.CustomQueue.itemProcessed(CustomQueue.java:302)
    at com.mycompany.collections.CustomQueue.trackCompleted(CustomQueue.java:147)
    at java.util.concurrent.ThreadPoolExecutor.afterExecute(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at java.lang.Thread.run(Unknown Source)

   Locked ownable synchronizers:
    - <0x00002aaf5f9ed518> (a java.util.concurrent.ThreadPoolExecutor$Worker)

"xxx>Job Read-1" daemon prio=10 tid=0x00002aca00183000 nid=0x6a39 waiting for monitor entry [0x0000000052c42000]
   java.lang.Thread.State: BLOCKED (on object monitor)
    at com.mycompany.collections.CustomQueue.remove(CustomQueue.java:101)
    - waiting to lock <0x00002aae6465a650> (a java.util.ArrayDeque)
    at com.mycompany.collections.CustomQueue.trim(CustomQueue.java:318)
    at com.mycompany.collections.CustomQueue.itemProcessed(CustomQueue.java:302)
    at com.mycompany.collections.CustomQueue.trackCompleted(CustomQueue.java:147)
    at java.util.concurrent.ThreadPoolExecutor.afterExecute(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at java.lang.Thread.run(Unknown Source)

   Locked ownable synchronizers:
    - <0x00002aaf5f9ecde8> (a java.util.concurrent.ThreadPoolExecutor$Worker)


"xxx>Job Read-0" daemon prio=10 tid=0x0000000006a83000 nid=0x6a36 waiting for monitor entry [0x000000005287f000]
   java.lang.Thread.State: BLOCKED (on object monitor)
        at com.mycompany.collections.CustomQueue.remove(CustomQueue.java:101)
    - waiting to lock <0x00002aae6465a650> (a java.util.ArrayDeque)
    at com.mycompany.collections.CustomQueue.trim(CustomQueue.java:318)
    at com.mycompany.collections.CustomQueue.itemProcessed(CustomQueue.java:302)
    at com.mycompany.collections.CustomQueue.trackCompleted(CustomQueue.java:147)
    at java.util.concurrent.ThreadPoolExecutor.afterExecute(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at java.lang.Thread.run(Unknown Source)
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这是提取的Java代码,它显示了错误的位置......

public class Deadlock {
        final Deque<Object> delegate  = new ArrayDeque<>();
        final long maxSize = Long.MAX_VALUE;

        private final AtomicLong totalExec = new AtomicLong();
        private final Map<Object, AtomicLong> totals = new HashMap<>();
        private final Map<Object, Deque<Long>> execTimes = new HashMap<>();

        public void trim() {
            //Possible optimization is evicting in chunks, segmenting by arrival time
            while (this.totalExec.longValue() > this.maxSize) {
                final Object t = this.delegate.peek();
                final Deque<Long> execTime = this.execTimes.get(t);
                final Long exec = execTime.peek();
                if (exec != null && this.totalExec.longValue() - exec > this.maxSize) {
                    //If Job Started Inside of Window, remove and re-loop
                    remove();
                }
                else {
                    //Otherwise exit the loop
                    break;
                }
            }
        }

        public Object remove() {
            Object removed;
            synchronized (this.delegate) { //4 Threads deadlocking on this line !
                removed = this.delegate.pollFirst();
            }
            if (removed != null) {
                itemRemoved(removed);
            }
            return removed;
        }

        public void itemRemoved(final Object t) {
            //Decrement Total & Queue
            final AtomicLong catTotal = this.totals.get(t);
            if (catTotal != null) {
                if (!this.execTimes.get(t).isEmpty()) {
                    final Long exec = this.execTimes.get(t).pollFirst();
                    if (exec != null) {
                        catTotal.addAndGet(-exec);
                        this.totalExec.addAndGet(-exec);
                    }
                }
            }
        }
    }
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Bor*_*der 5

文档中HashMap

请注意,此实现不同步.如果多个线程同时访问哈希映射,并且至少有一个线程在结构上修改了映射,则必须在外部进行同步.

(强调他们的)

您正在Map以不同步的方式读取和写入s.

我认为没有理由认为您的代码是线程安全的.

我建议你trim由于缺乏线程安全而导致无限循环.

输入同步块相对较慢,因此线程转储可能总是至少显示等待获取锁的几个线程.

  • @JohnVint代码允许修改.如果发生修改,则代码变得不可预测.这段代码不是线程安全的 - 进一步分析是没有意义的. (2认同)