读Lock-Free论文实践

论文地址:implementing Lock-Free Queue

论文大体讲的意思是:Lock-Base的程序的performance不好,并且a process inside the critical section can delay all operations inde nitely;所以基于以上的弊端,他们提出了Non-Blocking的算法,也就是CSW和FAA,当然就是CAS,而CAS也有最难以handler的情况,也就是ABA问题,他们给出了solution,也就是检查引用;他们分别给出了链表场景和数组场景的algorithm,最后是性能分析。

这篇笔记主要为了给Lock-Free提供一些实现方法和思路。

 

我们先看链表的情况:

 

 

 

双端链表,入队Enqueue方法是tail移动,出队Dequeue是Head移动,下面记录一下伪代码。

Enqueue(x)
q new record
q^:value x
q^:next NULL
repeat
p tail
succ Compare&Swap(p^:next, NULL, q)
if succ 6= TRUE
Compare&Swap(tail ; p; p^:next)
until succ = TRUE
Compare&Swap(tail ; p; q)
end
Dequeue() repeat p head
if p^:next = NULL error queue empty until Compare&Swap(head ; p; p^:next) return p^:next^:value end

Enqueue的思路是,先设置tail的next指针,如果成功,则把tail指针移动;Dequeue的思路是,如果head的copy p的next不为空,则进行移动,并在成功之后返回之前p的next的值,Java没有指针操作,只有使用unSafe类进行内存操作。(数组实现要比链表实现稳定多了)在跑线程的时候,add会有几率出现Runnable的情况,目前还不知道什么原因,最初的原因是把变量赋值写在for(;;)里,导致它一直不取值,只做循环体,将赋值写在循环体里好了一些,但还是会出现死循环问题。

 

import sun.misc.Unsafe;

import java.lang.reflect.Field;

/**
 * Created by MacBook on 2019/4/14.
 */
public class MyLockFreeLinkQueue<E> implements MyQueue<E>{

    Node<E> head;
    Node<E> tail;

    static Unsafe unsafe;

    private static final long headOffset;
    private static final long tailOffset;
    private static final long nextOffset;


    static{
        try{
            Field singleoneInstanceField = Unsafe.class.getDeclaredField("theUnsafe");

            singleoneInstanceField.setAccessible(true);

            unsafe = (Unsafe)singleoneInstanceField.get(null);

            headOffset = unsafe.objectFieldOffset
                    (MyLockFreeLinkQueue.class.getDeclaredField("head"));
            tailOffset = unsafe.objectFieldOffset
                    (MyLockFreeLinkQueue.class.getDeclaredField("tail"));
            nextOffset = unsafe.objectFieldOffset
                    (MyLockFreeLinkQueue.Node.class.getDeclaredField("next"));
        }catch (Exception e){
            throw new Error(e);
        }
    }


    static class Node<E>{
        E data;
        Node<E> next;

        public Node(E data, Node<E> next) {
            this.data = data;
            this.next = next;
        }

        public E getData() {
            return data;
        }

        public void setData(E data) {
            this.data = data;
        }

        public Node<E> getNext() {
            return next;
        }

        public void setNext(Node<E> next) {
            this.next = next;
        }
    }


    public MyLockFreeLinkQueue() {
        head = tail = new Node<>(null,null);
    }

    /**
     *
     Enqueue(x)
     q new record
     q^:value x
     q^:next NULL
     repeat
     p tail
     succ Compare&Swap(p^:next, NULL, q)
     if succ 6= TRUE
     Compare&Swap(tail ; p; p^:next)
     until succ = TRUE
     Compare&Swap(tail ; p; q)
     end
     * @param e
     * @return
     */
    @Override
    public boolean add(E e) {
        Node<E> q = new Node<>(e,null);
        for(;;){
            Node<E> p = tail;
            if(unsafe.compareAndSwapObject(p,nextOffset,null,q)){
                while(unsafe.compareAndSwapObject(this,tailOffset,p,q));
                break;
            }
        }
        return true;
    }


    /**
     *
     Dequeue()
     repeat
     p head
     if p^:next = NULL
     error queue empty
     until Compare&Swap(head ; p; p^:next)
     return p^:next^:value
     end
     * @return
     */
    @Override
    public E take() {
        for(;;){
            Node<E> p = head,next = p.getNext();
            if(next == null){
                return null;
            }else if(unsafe.compareAndSwapObject(this,headOffset,p,next)){
                p.setNext(null);// help gc
                return next.getData();
            }
        }
    }

}

 

 

 

这里我跑了2个生产者线程,2个消费者线程,数据有序的被消费了。

...
pool-1-thread-3 send [62] to queue; total 193 pool-1-thread-3 send [97] to queue; total 194 pool-1-thread-3 send [25] to queue; total 195 pool-1-thread-3 send [17] to queue; total 196 pool-1-thread-3 send [50] to queue; total 197 pool-1-thread-3 send [72] to queue; total 198 pool-1-thread-3 send [46] to queue; total 199 pool-1-thread-3 send [83] to queue; total 200 pool-1-thread-4 consumer [62],count 2n+1 result :125; total 193 pool-1-thread-4 consumer [97],count 2n+1 result :195; total 194 pool-1-thread-4 consumer [25],count 2n+1 result :51; total 195 pool-1-thread-4 consumer [17],count 2n+1 result :35; total 196 pool-1-thread-4 consumer [50],count 2n+1 result :101; total 197 pool-1-thread-4 consumer [72],count 2n+1 result :145; total 198 pool-1-thread-4 consumer [46],count 2n+1 result :93; total 199 pool-1-thread-4 consumer [83],count 2n+1 result :167; total 200

 

接下来,我实践了环形数组的实现,基于我之前实现的BlockingQueue,这个Lock-Free Queue会变得比较简单。

 

 

 

import java.util.concurrent.atomic.AtomicInteger;

/**
 * Created by MacBook on 2019/4/13.
 */
public class MyLockFreeQueue<E> implements MyQueue<E>{
    private Object[] data;
    private AtomicInteger takeIndex;
    private AtomicInteger putIndex;
    private AtomicInteger size;
    private static final int DEFAULT_CAPACITY = 10;

    public MyLockFreeQueue (){
        this(DEFAULT_CAPACITY);
    }
    public MyLockFreeQueue(int initCapacity){
        if(initCapacity < 0){
            throw new IllegalStateException("initCapacity must not be negative");
        }
        data = new Object[initCapacity];
        takeIndex = new AtomicInteger(0);
        putIndex = new AtomicInteger(0);
        size = new AtomicInteger(0);
    }

    public boolean add(E e){
        if(e == null){
            throw new NullPointerException("the element you put can't be null");
        }
        for(int index = putIndex.get();;){
            if(size.get() == data.length){
                return false;
            }
            int expect = (index == data.length - 1)?0:(index+1);
            if(putIndex.compareAndSet(index,expect)){
                data[index] = e;
                size.incrementAndGet();
                return true;
            }
        }
    }
    public E take(){
        for(int index = takeIndex.get();;){
            if(size.get() == 0){
                return null;
            }
            int expect = (index == data.length - 1)?0:(index+1);
            E e = (E)data[index];
            if(takeIndex.compareAndSet(index,expect)){
                size.decrementAndGet();
                return e;
            }
        }
    }
}

思路就是,使用两个标记入队和出队的Atom Integer对象,在成功申请当前格子之后,给当前格子赋值,使用size来判断是否EMPTY和FULL。这里依然有一点缺陷,就是index和size不同步的问题,不过我也是跑了2+2线程,也是有序消费了。

...pool-1-thread-3 send [81] to queue; total 188
pool-1-thread-2 consumer [81],count 2n+1 result :163; total 188
pool-1-thread-3 send [1] to queue; total 189
pool-1-thread-2 consumer [1],count 2n+1 result :3; total 189
pool-1-thread-2 consumer [19],count 2n+1 result :39; total 190
pool-1-thread-3 send [19] to queue; total 190
pool-1-thread-3 send [61] to queue; total 191
pool-1-thread-2 consumer [61],count 2n+1 result :123; total 191
pool-1-thread-3 send [16] to queue; total 192
pool-1-thread-2 consumer [16],count 2n+1 result :33; total 192
pool-1-thread-3 send [74] to queue; total 193
pool-1-thread-2 consumer [74],count 2n+1 result :149; total 193
pool-1-thread-3 send [38] to queue; total 194
pool-1-thread-2 consumer [38],count 2n+1 result :77; total 194
pool-1-thread-3 send [32] to queue; total 195
pool-1-thread-2 consumer [32],count 2n+1 result :65; total 195
pool-1-thread-3 send [9] to queue; total 196
pool-1-thread-2 consumer [9],count 2n+1 result :19; total 196
pool-1-thread-3 send [77] to queue; total 197
pool-1-thread-2 consumer [77],count 2n+1 result :155; total 197
pool-1-thread-3 send [69] to queue; total 198
pool-1-thread-2 consumer [69],count 2n+1 result :139; total 198
pool-1-thread-3 send [52] to queue; total 199
pool-1-thread-2 consumer [52],count 2n+1 result :105; total 199
pool-1-thread-3 send [81] to queue; total 200
pool-1-thread-2 consumer [81],count 2n+1 result :163; total 200

 

        ExecutorService executor = Executors.newFixedThreadPool(6);
        MyLockFreeQueue<Integer> queue = new MyLockFreeQueue();
        Worker<Integer> pro = new Provider(queue);
        Worker<Integer> con = new Consumer(queue);

        executor.submit(pro);
        executor.submit(con);
        executor.submit(pro);
        executor.submit(con);
        executor.submit(pro);
        executor.submit(con);
        
        executor.shutdown();

 

posted @ 2019-04-14 14:58  天目山电鳗  阅读(208)  评论(0编辑  收藏