一尘不染

使用`LinkedBlockingQueue`可能导致空指针异常

java

我最近正在学习Java并发编程。我知道final关键字可以保证安全的发布。但是,当我阅读LinkedBlockingQueue源代码时,发现headand
last字段未使用final关键字。我发现该enqueue方法在方法中被调用put,并且该enqueue方法直接将值分配给last.next。此时,last可能是null因为last未使用声明final。我的理解正确吗?虽然lock可以保证last读写线程的安全性,但是可以lock
保证last不是一个正确的初始值null

public class LinkedBlockingQueue<E> extends AbstractQueue<E>
        implements BlockingQueue<E>, java.io.Serializable {
transient Node<E> head;
private transient Node<E> last;
public LinkedBlockingQueue(int capacity) {
        if (capacity <= 0) throw new IllegalArgumentException();
        this.capacity = capacity;
        last = head = new Node<E>(null);
    }
 private void enqueue(Node<E> node) {
        // assert putLock.isHeldByCurrentThread();
        // assert last.next == null;
        last = last.next = node;
    }

public void put(E e) throws InterruptedException {
        if (e == null) throw new NullPointerException();
        // Note: convention in all put/take/etc is to preset local var
        // holding count negative to indicate failure unless set.
        int c = -1;
        Node<E> node = new Node<E>(e);
        final ReentrantLock putLock = this.putLock;
        final AtomicInteger count = this.count;
        putLock.lockInterruptibly();
        try {
            /*
             * Note that count is used in wait guard even though it is
             * not protected by lock. This works because count can
             * only decrease at this point (all other puts are shut
             * out by lock), and we (or some other waiting put) are
             * signalled if it ever changes from capacity. Similarly
             * for all other uses of count in other wait guards.
             */
            while (count.get() == capacity) {
                notFull.await();
            }
            enqueue(node);
            c = count.getAndIncrement();
            if (c + 1 < capacity)
                notFull.signal();
        } finally {
            putLock.unlock();
        }
        if (c == 0)
            signalNotEmpty();
    }
}

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2020-12-03

共1个答案

一尘不染

根据此博客文章https://shipilev.net/blog/2014/safe-public-
construction/甚至final在构造函数中写入一个属性也足以实现安全的初始化(因此,您的对象将始终被安全地发布)。并且capacity属性声明为final

简而言之,我们在以下三种情况下会产生障碍:

最后一个领域是这样写的。注意,我们并不关心实际写入哪个字段,我们在退出(initializer)方法之前无条件地发出了屏障。这意味着,如果您至少有一个final字段写操作,则final字段语义将扩展到构造函数中编写的所有其他字段。

2020-12-03