LinkedBlockingQueue源码解析

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1. 介绍

​ 可选的有界的阻塞队列,基于链表实现,FIFO,默认长度Integer.MAX_VALUE,超过会阻塞。内部用两把锁实现,takeLock,putLock,避免了读和写相互竞争的情况。但是在删除等遍历操作时,会锁两把锁。

2. 源码

2.1 构造函数

 	public LinkedBlockingQueue() {
        this(Integer.MAX_VALUE);
    }
    public LinkedBlockingQueue(int capacity) {
        if (capacity <= 0) throw new IllegalArgumentException();
        this.capacity = capacity;
        last = head = new Node<E>(null);
    }
    public LinkedBlockingQueue(Collection<? extends E> c) {
        this(Integer.MAX_VALUE);
        final ReentrantLock putLock = this.putLock;
        putLock.lock(); // Never contended, but necessary for visibility
        try {
            int n = 0;
            for (E e : c) {
                if (e == null)
                    throw new NullPointerException();
                if (n == capacity)
                    throw new IllegalStateException("Queue full");
                enqueue(new Node<E>(e));
                ++n;
            }
            count.set(n);
        } finally {
            putLock.unlock();
        }
    

2.2 put

​ 若队列满,阻塞

 public void put(E e) throws InterruptedException {
        if (e == null) throw new NullPointerException();
        int c = -1;
        Node<E> node = new Node<E>(e);
        final ReentrantLock putLock = this.putLock;
        final AtomicInteger count = this.count;
        putLock.lockInterruptibly();
        try {
            while (count.get() == capacity) {
                notFull.await();
            }
            enqueue(node);
            c = count.getAndIncrement();
            if (c + 1 < capacity)
                notFull.signal();
        } finally {
            putLock.unlock();
        }
        if (c == 0)
            signalNotEmpty();
    }
    private void enqueue(Node<E> node) {
        // assert putLock.isHeldByCurrentThread();
        // assert last.next == null;
        last = last.next = node;
    }
  1. 获取putlock锁
  2. 如果容量满,等待
  3. 入队,数量+1
  4. 若容量小于最大值,唤醒处于插入等待的阻塞线程
  5. 如果原来容量为0,唤醒处于获取等待的阻塞线程

2.3 offer

​ 插入元素,指定最大等待时间,超时未插入,返回false

    public boolean offer(E e, long timeout, TimeUnit unit)
        throws InterruptedException {

        if (e == null) throw new NullPointerException();
        long nanos = unit.toNanos(timeout);
        int c = -1;
        final ReentrantLock putLock = this.putLock;
        final AtomicInteger count = this.count;
        putLock.lockInterruptibly();
        try {
            while (count.get() == capacity) {
                if (nanos <= 0)
                    return false;
                nanos = notFull.awaitNanos(nanos);
            }
            enqueue(new Node<E>(e));
            c = count.getAndIncrement();
            if (c + 1 < capacity)
                notFull.signal();
        } finally {
            putLock.unlock();
        }
        if (c == 0)
            signalNotEmpty();
        return true;
    }

非阻塞,插入元素,容量已满,直接返回false

    public boolean offer(E e) {
        if (e == null) throw new NullPointerException();
        final AtomicInteger count = this.count;
        if (count.get() == capacity)
            return false;
        int c = -1;
        Node<E> node = new Node<E>(e);
        final ReentrantLock putLock = this.putLock;
        putLock.lock();
        try {
            if (count.get() < capacity) {
                enqueue(node);
                c = count.getAndIncrement();
                if (c + 1 < capacity)
                    notFull.signal();
            }
        } finally {
            putLock.unlock();
        }
        if (c == 0)
            signalNotEmpty();
        return c >= 0;
    }

2.4 take

阻塞式获取,直到有元素插入进来

    public E take() throws InterruptedException {
        E x;
        int c = -1;
        final AtomicInteger count = this.count;
        final ReentrantLock takeLock = this.takeLock;
        takeLock.lockInterruptibly();
        try {
            while (count.get() == 0) {
                notEmpty.await();
            }
            x = dequeue();
            c = count.getAndDecrement();
            if (c > 1)
                notEmpty.signal();
        } finally {
            takeLock.unlock();
        }
        if (c == capacity)
            signalNotFull();
        return x;
    }
  1. 获取takelock锁
  2. 若队列为空,等待
  3. 出队,数量-1
  4. 若容量大于1,唤醒处于获取等待的阻塞线程
  5. 若原来容量已经满了(take后就不满了),唤醒处于插入等待的阻塞线程

2.5 pull

获取元素,有最大等待时间,超时返回null

 	public E poll(long timeout, TimeUnit unit) throws InterruptedException {
        E x = null;
        int c = -1;
        long nanos = unit.toNanos(timeout);
        final AtomicInteger count = this.count;
        final ReentrantLock takeLock = this.takeLock;
        takeLock.lockInterruptibly();
        try {
            while (count.get() == 0) {
                if (nanos <= 0)
                    return null;
                nanos = notEmpty.awaitNanos(nanos);
            }
            x = dequeue();
            c = count.getAndDecrement();
            if (c > 1)
                notEmpty.signal();
        } finally {
            takeLock.unlock();
        }
        if (c == capacity)
            signalNotFull();
        return x;
    }

非阻塞获取元素,若队列为空,返回null

	 public E poll() {
        final AtomicInteger count = this.count;
        if (count.get() == 0)
            return null;
        E x = null;
        int c = -1;
        final ReentrantLock takeLock = this.takeLock;
        takeLock.lock();
        try {
            if (count.get() > 0) {
                x = dequeue();
                c = count.getAndDecrement();
                if (c > 1)
                    notEmpty.signal();
            }
        } finally {
            takeLock.unlock();
        }
        if (c == capacity)
            signalNotFull();
        return x;
    }