Question

所以，我的Phaser非常灵活，但似乎我错过了一些东西。我已经成功使用了CyclicBarrier，但现在我也想要一些更灵活的东西。所以这是代码：

声明：

private static final CountDownLatch synchronizer = new CountDownLatch(1);
private static AtomicBoolean HAS_TIMED_OUT = new AtomicBoolean(false);

的代码：

try {
    logger.INFO("CONNECTED - Peer ID properties: " + SYS_NEWLINE + peerSocket + SYS_NEWLINE + pID, true);

    final int peerKQueries = sp.getInteger(peerSocket);
    peerObjects = new String[peerKQueries];
    peerValues = new BigDecimal[peerKQueries];
    for ( int i = 0; i < peerObjects.length; i++ )
       peerObjects[i] = sp.getString(peerSocket);
    for ( int i = 0; i < peerValues.length; i++ )
       peerValues[i] = sp.getBigDecimal(peerSocket);

    final int phase1a = htPhaser1a.arrive();
    if ( phase1a < 0 ) {
        logger.ERROR("Rejecting Super Peer thread " + THREAD_ID + " because it arrived lately for Phase 1a!", true);
        sp.close(peerSocket);
        throw new IllegalThreadStateException();
    } else {
        logger.INFO(pID + " -> Arrived in HT phase 1a. Total arrivals: "+htPhaser1a.getArrivedParties(), true);
        logger.INFO("Super Peer thread " + THREAD_ID + " will advance to HT Phase 1b/2 (phase number is "+phase1a+").", true);
        // The last peer should also unblock the barrier.
        if ( htPhaser1a.getArrivedParties() == TOTAL_PEERS.get() ) {
          htPhaser1a.arrive();
          synchronizer.countDown();
        }
            htPhaser1a.awaitAdvanceInterruptibly(phase1a, 30, TimeUnit.SECONDS);
    }

} catch (IOException e) {
    logger.ERROR("Super Peer thread " + THREAD_ID + " encountered an I/O error.", true);
    sp.close(peerSocket);
    throw new IllegalThreadStateException();
} catch (TimeoutException e) {
    logger.INFO("Super Peer thread " + THREAD_ID + " timed out but will advance to HT Phase 1b/2.", true);
    if ( HAS_TIMED_OUT.compareAndSet(false, true) ) {
        logger.INFO("Parties NOT arrived in the timeout: "+(htPhaser1a.getUnarrivedParties()-1), true);
        resetCriticalData(htPhaser1a.getArrivedParties());
        htPhaser1a.forceTermination();
        instantiateHTPhase1b();
        instantiateHTPhase2();
        instantiateHTPatch();
        synchronizer.countDown();
    }
} finally {
    logger.INFO("Super Peer thread "+THREAD_ID+" is blocked!", true);
    synchronizer.await();
    logger.INFO("Super Peer thread's "+THREAD_ID+" blocking waived!", true);
}

sp.getSomething();是I / O电话。
考虑到这个代码示例是由多个线程运行的。

这是我的问题：我确保不会超过MAX_CLIENTS到达移相器，所以如果MAX_CLIENTS到达一切都很好。但是，我遇到了TimeoutException的问题。第一个是时间窗口（又称竞争条件），客户端（比如线程A）将能够到达阶段，然后在线程B中发生TimeoutException，我动态地实例化线程B中的另一个移相器，其中包含到达方的数量（比如说5），但是线程A已经到达阶段（又名phase1a未被发现为＆lt; 0）。我怎么能纠正这个？我正在考虑使用信号量，但我认为这不值得付出努力，因为我可能需要重新考虑我这样做的方式。我还考虑过使用Timer并递增AtomicInteger变量，并在计时器到期时动态实例化Phaser。有关如何处理此问题的任何想法？

编辑：
文档有一个bulkRegister(int parties)方法，但有点奇怪的措辞：

在此移相器中添加给定数量的新未获得的方。如果正在进行的onAdvance（int，int）调用正在进行中，则此方法可能在返回之前等待其完成。如果此移相器具有父级，并且给定的参与方数大于零，并且此移相器之前没有注册方，则此子移相器也向其父级注册。如果此移相器终止，则注册尝试无效，并返回负值。

问题：“可能”这个词让我困惑！ “愿意”或“可能”如意？ 修改：
解决。检查下面的答案。

Answer 1

声明：

private static final CountDownLatch PEER = new CountDownLatch(1);
private static AtomicBoolean HAS_TIMED_OUT = new AtomicBoolean(false);
htPeerPhaser = new Phaser();

的代码：

...
htPeerPhaser.register(); // Called only once.
...
// Note: Server application has guaranteed that no more than the maximum number of peers will arrive.
try {
    logger.INFO("CONNECTED - Peer ID properties: " + SYS_NEWLINE + peerSocket + SYS_NEWLINE + pID, true);
    final int peerKQueries = sp.getInteger(peerSocket);
    peerObjects = new String[peerKQueries];
    peerValues = new BigDecimal[peerKQueries];
    for ( int i = 0; i < peerObjects.length; i++ )
        peerObjects[i] = sp.getString(peerSocket);
    for ( int i = 0; i < peerValues.length; i++ )
        peerValues[i] = sp.getBigDecimal(peerSocket);
    final int registrationID = htPeerPhaser.bulkRegister(1);
    if ( registrationID < 0 ) {
        logger.ERROR("Rejecting Super Peer thread " + THREAD_ID + " because peer registration has stopped!", true);
        sp.close(peerSocket);
        throw new IllegalThreadStateException();
    }
    logger.INFO(pID + " -> Registered for HT phase 1.", true);
    logger.INFO("Super Peer thread " + THREAD_ID + " will advance to HT Phase 1/2.", true);
    // The last peer should also unblock the barrier.
    if ( htPeerPhaser.getRegisteredParties() == TOTAL_PEERS.get()+1 ) {
        htPeerPhaser.forceTermination();
        PEER.countDown();
    }
    htPeerPhaser.awaitAdvanceInterruptibly(registrationID, 30, TimeUnit.SECONDS);

} catch (IOException e) {
    logger.ERROR("Super Peer thread " + THREAD_ID + " encountered an I/O error.", true);
    sp.close(peerSocket);
    throw new IllegalThreadStateException();
} catch (TimeoutException e) {
    htPeerPhaser.forceTermination();
    logger.INFO("Super Peer thread " + THREAD_ID + " timed out but will advance to HT Phase 1b/2.", true);
    if ( HAS_TIMED_OUT.compareAndSet(false, true) && htPeerPhaser.getRegisteredParties() < TOTAL_PEERS.get()+1 ) {
        final int arrivedPeers = htPeerPhaser.getRegisteredParties()-1;
        logger.INFO("Parties that arrived before timeout: "+arrivedPeers, true);
        final int unarrivedPeers = TOTAL_PEERS.get()-arrivedPeers;
        logger.INFO("Parties NOT arrived due to timeout: "+unarrivedPeers, true);
        resetCriticalData(arrivedPeers);
        instantiateHTPhase1b();
        instantiateHTPhase2();
        instantiateHTPatch();
        PEER.countDown();
        logger.INFO("Super Peer thread " + THREAD_ID + " re-instantiated critical data.", true);
    }
}
logger.INFO("Super Peer thread "+THREAD_ID+" is blocked!", true);
PEER.await();
logger.INFO("Super Peer thread's "+THREAD_ID+" blocking waived!", true);

Phaser问题（使用JSR166y和版本6的最新JDK）

1 个答案: