Redis中Redisson红锁(Redlock)使用原理

RLock lock1 = redissonInstance1.getLock("lock1");
RLock lock2 = redissonInstance2.getLock("lock2");
RLock lock3 = redissonInstance3.getLock("lock3");
 
RedissonRedLock lock = new RedissonRedLock(lock1, lock2, lock3);
// 同时加锁：lock1 lock2 lock3
// 红锁在大部分节点上加锁成功就算成功。
lock.lock();
...
lock.unlock();

RedissonRedLock lock = new RedissonRedLock(lock1, lock2, lock3);
// 给lock1，lock2，lock3加锁，如果没有手动解开的话，10秒钟后将会自动解开
lock.lock(10, TimeUnit.SECONDS);
 
// 为加锁等待100秒时间，并在加锁成功10秒钟后自动解开
boolean res = lock.tryLock(100, 10, TimeUnit.SECONDS);
...
lock.unlock();

final List<RLock> locks = new ArrayList<>();
 
/**
 * Creates instance with multiple {@link RLock} objects.
 * Each RLock object could be created by own Redisson instance.
 *
 * @param locks - array of locks
 */
public RedissonMultiLock(RLock... locks) {
    if (locks.length == 0) {
        throw new IllegalArgumentException("Lock objects are not defined");
    }
    this.locks.addAll(Arrays.asList(locks));
}
 
public boolean tryLock(long waitTime, long leaseTime, TimeUnit unit) throws InterruptedException {
    long newLeaseTime = -1;
    if (leaseTime != -1) {
        newLeaseTime = unit.toMillis(waitTime)*2;
    }
    
    long time = System.currentTimeMillis();
    long remainTime = -1;
    if (waitTime != -1) {
        remainTime = unit.toMillis(waitTime);
    }
    long lockWaitTime = calcLockWaitTime(remainTime);
    /**
     * 1. 允许加锁失败节点个数限制（N-(N/2+1)）
     */
    int failedLocksLimit = failedLocksLimit();
    /**
     * 2. 遍历所有节点通过EVAL命令执行lua加锁
     */
    List<RLock> acquiredLocks = new ArrayList<>(locks.size());
    for (ListIterator<RLock> iterator = locks.listIterator(); iterator.hasNext();) {
        RLock lock = iterator.next();
        boolean lockAcquired;
        /**
         *  3.对节点尝试加锁
         */
        try {
            if (waitTime == -1 && leaseTime == -1) {
                lockAcquired = lock.tryLock();
            } else {
                long awaitTime = Math.min(lockWaitTime, remainTime);
                lockAcquired = lock.tryLock(awaitTime, newLeaseTime, TimeUnit.MILLISECONDS);
            }
        } catch (RedisResponseTimeoutException e) {
            // 如果抛出这类异常，为了防止加锁成功，但是响应失败，需要解锁所有节点
            unlockInner(Arrays.asList(lock));
            lockAcquired = false;
        } catch (Exception e) {
            // 抛出异常表示获取锁失败
            lockAcquired = false;
        }
        
        if (lockAcquired) {
            /**
             *4. 如果获取到锁则添加到已获取锁集合中
             */
            acquiredLocks.add(lock);
        } else {
            /**
             * 5. 计算已经申请锁失败的节点是否已经到达 允许加锁失败节点个数限制 （N-(N/2+1)）
             * 如果已经到达， 就认定最终申请锁失败，则没有必要继续从后面的节点申请了
             * 因为 Redlock 算法要求至少N/2+1 个节点都加锁成功，才算最终的锁申请成功
             */
            if (locks.size() - acquiredLocks.size() == failedLocksLimit()) {
                break;
            }
 
            if (failedLocksLimit == 0) {
                unlockInner(acquiredLocks);
                if (waitTime == -1 && leaseTime == -1) {
                    return false;
                }
                failedLocksLimit = failedLocksLimit();
                acquiredLocks.clear();
                // reset iterator
                while (iterator.hasPrevious()) {
                    iterator.previous();
                }
            } else {
                failedLocksLimit--;
            }
        }
 
        /**
         * 6.计算 目前从各个节点获取锁已经消耗的总时间，如果已经等于最大等待时间，则认定最终申请锁失败，返回false
         */
        if (remainTime != -1) {
            remainTime -= System.currentTimeMillis() - time;
            time = System.currentTimeMillis();
            if (remainTime <= 0) {
                unlockInner(acquiredLocks);
                return false;
            }
        }
    }
 
    if (leaseTime != -1) {
        List<RFuture<Boolean>> futures = new ArrayList<>(acquiredLocks.size());
        for (RLock rLock : acquiredLocks) {
            RFuture<Boolean> future = ((RedissonLock) rLock).expireAsync(unit.toMillis(leaseTime), TimeUnit.MILLISECONDS);
            futures.add(future);
        }
        
        for (RFuture<Boolean> rFuture : futures) {
            rFuture.syncUninterruptibly();
        }
    }
 
    /**
     * 7.如果逻辑正常执行完则认为最终申请锁成功，返回true
     */
    return true;
}