多线程（十、AQS原理-ReentrantLock公平锁）

ReentrantLock介绍

ReentrantLock 基于AQS实现了公平和非公平的独占锁功能。

ReentrantLock定义AQS的同步状态（synchronization state）如下：

State为0表示锁可用；为1表示被占用；为N表示锁重入的次数，是独占资源。

ReentrantLock实现公平锁原理

案例代码如下：

1、启动文件

public class Main {

    public static void main(String[] args) throws ParseException {

        ReentrantLock lock = new ReentrantLock(true);
        Thread t1 = new Thread(new Task(lock),"Thread-1");
        Thread t2 = new Thread(new Task(lock),"Thread-2");
        Thread t3 = new Thread(new Task(lock),"Thread-3");

        t1.start();
        t2.start();
        t3.start();
    }
}

2、Task

import java.util.concurrent.locks.ReentrantLock;

public class Task implements Runnable{

    private ReentrantLock lock;

    public Task(ReentrantLock lock) {
        this.lock = lock;
    }

    @Override
    public void run() {

        try {
            lock.lock();
            System.out.println(Thread.currentThread().getName() + "获取到锁....");
            Thread.sleep(2000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            System.out.println(Thread.currentThread().getName() + "释放锁....");
            lock.unlock();
        }
    }
}

3、执行结果：

案例分析

1、Thread-1调用lock方法

ReentrantLock内部继承AQS实现了1个抽象类Sync

继承于Sync实现了2个内部类FairSync（公平的）和NonfairSync（非公平的）

此时调用FairSync的lock方法

acquire方法来自AQS，注意参数是1

tryAcquire是ReentrantLock自己实现的，尝试获取锁

protected final boolean tryAcquire(int acquires) { //参数是1
        final Thread current = Thread.currentThread(); //当前线程
        int c = getState(); //获取当前同步状态
        if (c == 0) { //如果是0，则锁没有被占用
            //等待队列中，前面没有其他等待的线程，则用CAS的方法更新同步状态state
            if (!hasQueuedPredecessors() &&
                    compareAndSetState(0, acquires)) {
                setExclusiveOwnerThread(current); //成功的话，则设置锁的占有线程为当前线程
                return true; //返回获取资源成功
            }
        }
        //如果锁已经被占用，则判断是不是自己占用的
        else if (current == getExclusiveOwnerThread()) {
            int nextc = c + acquires;//如果是自己占用的，则是重入，增加state值，累加1
            if (nextc < 0) //重入次数过大，抛出异常
                throw new Error("Maximum lock count exceeded");
            setState(nextc); //设置state值
            return true; //重入返回ture
        }
        return false;//没有获取资源返回false
    }

Thread-1获取了锁资源，没有释放。

2、Thread-2，开始请求资源，调用lock，此时锁资源还被Thread-1占用

addWaiter方法：

private Node addWaiter(Node mode) {
        //把当前线程包装成节点，准备放入等待队列
        Node node = new Node(Thread.currentThread(), mode);
        // Try the fast path of enq; backup to full enq on failure
        //尝试直接把节点设置成队尾，否则执行enq
        Node pred = tail;
        if (pred != null) {
            node.prev = pred;//当前节点的上一个节点是之前的队尾节点
            if (compareAndSetTail(pred, node)) {
                pred.next = node;
                return node;
            }
        }
        //当前节点插入队尾
        enq(node);
        return node;
    }

enq自旋+初始化等待队列，并返回Thread-2节点

private Node enq(final Node node) {
        //采用自旋，保证节点插入
        for (;;) {
            Node t = tail;
            if (t == null) { // Must initialize 如果队列为空，则创建一个空的节点，设置为头尾节点
                if (compareAndSetHead(new Node()))
                    tail = head;
            } else {
                node.prev = t;
                if (compareAndSetTail(t, node)) { //队列不为空，追加到队尾
                    t.next = node;
                    return t;
                }
            }
        }
    }

然后对Thread-2包装节点执行acquireQueued

final boolean acquireQueued(final Node node, int arg) {
        boolean failed = true;
        try {
            boolean interrupted = false;
            for (;;) {
                //判断节点的前任节点是不是头节点，头节点是一个空节点
                final Node p = node.predecessor();
                //如果是头节点，则说明当前节点是队列里的第一个节点，首节点。
                //则尝试获取锁资源，此处因为Thread-1占用着资源，则失败
                if (p == head && tryAcquire(arg)) {
                    setHead(node);
                    p.next = null; // help GC
                    failed = false;
                    return interrupted;
                }
                //失败之后，则判断当前节点线程Thread-2是不是可以阻塞
                if (shouldParkAfterFailedAcquire(p, node) &&
                        parkAndCheckInterrupt())
                    interrupted = true;
            }
        } finally {
            if (failed)
                cancelAcquire(node);
        }
    }

是否阻塞shouldParkAfterFailedAcquire

private static boolean shouldParkAfterFailedAcquire(Node pred, Node node) {
        int ws = pred.waitStatus;//前驱节点的状态
        if (ws == Node.SIGNAL) //如果是SIGNAL，则说明前驱节点状态可以唤醒后继节点，可以阻塞
            /*
             * This node has already set status asking a release
             * to signal it, so it can safely park.
             */
            return true;
        if (ws > 0) {
            /*
             * Predecessor was cancelled. Skip over predecessors and
             * indicate retry.
             */
            do {
                node.prev = pred = pred.prev; //只有CANCELLED状态大于0，则把取消状态的节点从队列删除
            } while (pred.waitStatus > 0);
            pred.next = node;
        } else {
            /*
             * waitStatus must be 0 or PROPAGATE.  Indicate that we
             * need a signal, but don't park yet.  Caller will need to
             * retry to make sure it cannot acquire before parking.
             */
            compareAndSetWaitStatus(pred, ws, Node.SIGNAL);//设置前驱节点为SIGNAL状态
        }
        return false;
    }

如果可以阻塞，则调用parkAndCheckInterrupt，阻塞线程，至此Thread-2进入队列，并阻塞了，耐心等待

3、Thread-3同Thread-2，略过

4、Thread-1释放锁资源

release方法：

tryRelease

然后通过unparkSuccessor，唤醒首节点，保证公平策略。

至此Thread-1释放完成，Thread-2可以获得资源，依次类推。