并发双链表-多个生产者/消费者FIFO队列-死锁

Question

我正在处理MPMC FIFO队列的双向链接列表（主要用于演示）。 我现在的目标是试图纠正我的错误一段时间，但我并没有真正取得进展。

在所有Producer-Threads产生所有值后不久，我陷入了僵局。 我将问题归结为以下内容：

锁获取：

<THREAD ID> | <MESSAGE> | <MUTEX ADDRESS>

4460175360 : TRYLOCK HEAD:   0x7fb380c039d0
4460175360 : LOCKED HEAD:     0x7fb380c039d0
4460175360 : RELEASED HEAD: 0x7fb380c039d0

4460175360 : TRYLOCK HEAD:    0x7fb380c039d0
4460175360 : LOCKED HEAD:      0x7fb380c039d0
4460175360 : RELEASED HEAD:  0x7fb380c039d0

4460175360 : TRYLOCK HEAD:    0x7fb380c039d0
4460175360 : LOCKED HEAD:      0x7fb380c039d0
4460175360 : RELEASED HEAD:  0x7fb380c039d0

4460175360 : TRYLOCK HEAD:    0x7fb380c039d0
4459638784 : TRYLOCK TAIL:       0x7fb380c039d0
4460711936 : TRYLOCK TAIL:       0x7fb380c039d0
4460175360 : LOCKED HEAD:      0x7fb380c039d0
4460175360 : RELEASED HEAD:  0x7fb380c039d0

4460175360 : TRYLOCK HEAD: 0x7fb380c039d0   <---- THIS TRY-LOCK-HEAD NEVER GETS CONFIRMED!
4459638784 : LOCKED TAIL:  0x7fb380c039d0 
4459638784 : RELEASE MUTEX: 0x7fb380c039d0 <---- THE PRODUCER THREAD RELEASED 0x7fb380c039d0
Producer-Thread 4459638784 produced: 0

---- NOW ONE ELEMENT IN QUEUE ----

4460711936 : LOCKED TAIL:  0x7fb381800020 <---- ADDRESS OF LOCK HAS CHANGED WHICH IS FINE (because an element has been added)
4460711936 : RELEASE MUTEX AT ADDR: 0x7fb381800020
Producer-Thread 4460711936 produced: 0

---- NOW TWO ELEMENTS IN QUEUE ----

4460711936 : TRYLOCK TAIL: 0x7fb381a00020
4459638784 : TRYLOCK TAIL: 0x7fb381a00020
4460711936 : LOCKED TAIL:   0x7fb381a00020
4460711936 : RELEASE MUTEX AT ADDR: 0x7fb381a00020
Producer-Thread 4460711936 produced: 1

---- NOW THREE ELEMENTS IN QUEUE ----

4459638784 : LOCKED TAIL: 0x7fb380d00060  <---- AGAIN ADDRESS HAS CHANGED -- FINE
4459638784 : RELEASE MUTEX AT ADDR: 0x7fb380d00060
Producer-Thread 4459638784 produced: 1

---- NOW FOUR ELEMENTS IN QUEUE ---- PRODUCERS ARE DONE ----
---- CONSUMER THREAD BLOCKS - STILL TRYING TO LOCK 0x7fb380c039d0 ----

---- BUT NOBODY ELSE HOLDS 0x7fb380c039d0 ---- ALSO NO SELF-DEADLOCK?

GDB告诉我，两个生产者线程都已终止，并且唯一在addr 0x7fb380c039d0等待mutx的线程是线程2（唯一的使用者线程）：

    (gdb) info threads
* 2                         0x00007fff8edc5dfd in pthread_mutex_lock ()
  1 "com.apple.main-thread" 0x00007fff8e715386 in __semwait_signal ()
(gdb) bt
#0  0x00007fff8e715122 in __psynch_mutexwait ()
#1  0x00007fff8edc5dfd in pthread_mutex_lock ()
#2  0x00000001000011a8 in ConcurrentDoublyLinkedList<int>::consumeNode (this=0x1001000e0) at ConcurrentDList.h:142
#3  0x0000000100000bd4 in consumeValues (ctx=0x7fff5fbffa78) at ConcurrentDList.cpp:29
#4  0x00007fff8edc07a2 in _pthread_start ()
#5  0x00007fff8edad1e1 in thread_start ()
(gdb) f 2
#2  0x00000001000011a8 in ConcurrentDoublyLinkedList<int>::consumeNode (this=0x1001000e0) at ConcurrentDList.h:142
142             pthread_mutex_lock(&head_->mutex);
(gdb) info reg
rax            0x200012d    33554733
rbx            0x100281000  4297592832
rcx            0x100280da8  4297592232
rdx            0x100    256
rsi            0x403    1027
rdi            0x1001039f0  4296030704
rbp            0x100280ed0  0x100280ed0
rsp            0x100280e00  0x100280e00
r8             0x2060   8288
r9             0x100280720  4297590560
r10            0xf9d79  1023353
r11            0x206    518
r12            0x1303   4867
r13            0x0  0
r14            0x7fff5fbffa78   140734799805048
r15            0x100000bb0  4294970288
rip            0x1000011a8  0x1000011a8 <ConcurrentDoublyLinkedList<int>::consumeNode()+110>
eflags         0x206    518
cs             0x7  7
ss             0x0  0
ds             0x0  0
es             0x0  0
fs             0x0  0
gs             0x0  0
(gdb) p *(pthread_mutex_t*) 0x1001039f0
$34 = {
  __sig = 1297437784, 
  __opaque = "\000\000\000\000` \000\000\000\000\000\000\000\000\000\000\003\004\000\000\000\002\000\000{?\017\000\000\000\000\000\b:\020\000\001\000\000\000\f:\020\000\001\000\000\000\000\000\000\000\000\000\000"
}

不幸的是，我无法捕获pthread_mutex_t，因为它是不透明的类型。 （尽管我在GDB中启用了不透明类型？）。 否则，最好查看该互斥对象的所有者是谁。

下面列出了我的列表代码-对其进行了过多注释，以解释我对实现的看法。 请注意，到目前为止，这不是一个好的或高性能的实现，主要是出于说明目的。

CPP代码：

#include <stdlib.h>
#include <iostream>

#include <pthread.h>

template <class T>
class Node
{
    private:
       T value_;

       Node<T> *next_;
       Node<T> *prev_;

    public:

        pthread_mutex_t mutex;

        Node(T value, Node<T>* next, Node<T>* prev)
        {
            value_ = value;

            next_ = next;
            prev_ = prev;

            pthread_mutex_init(&mutex, NULL);
        }

        ~Node()
        {
            pthread_mutex_destroy(&mutex);
        }

        void setNext(Node<T>* next)
        {
            next_ = next;
        }        

        void setPrevious(Node<T>* prev)
        {
            prev_ = prev;
        }

        Node<T>* getNext()
        {
            return next_;
        } 

        Node<T>* getPrevious()
        {
            return prev_;
        }


        virtual bool isSentinel()
        {
            return false;
        } 
};

template <class T>
class SentinelNode : public Node<T>
{
    public:
      SentinelNode(T val, Node<T>* next, Node<T>* prev)
          : Node<T>(val, next, prev)
      {
      }

      virtual bool isSentinel()
      {
          return true;
      }
};

template <class T>
class ConcurrentDoublyLinkedList
{
    private:
        Node<T> *head_;
        Node<T> *tail_;

    public: 
        ConcurrentDoublyLinkedList()
        {
            head_ = tail_ = new SentinelNode<T>(NULL, NULL, NULL);
        }

        void produceNode(Node<T>* n)
        {  
            pthread_mutex_lock(&tail_->mutex);

            Node<T>* old = tail_;

            if(head_->isSentinel())
            {
                head_ = tail_ = n;
                pthread_mutex_unlock(&old->mutex);

                delete old;
                old = NULL;
            }
            else
            {
                n->setNext(tail_);
                tail_->setPrevious(n);
                tail_ = n;

                pthread_mutex_unlock(&old->mutex);
            }
        }

        Node<T>* consumeNode()
        {
            pthread_mutex_lock(&head_->mutex);

            if(head_->isSentinel())
            {
                 /* the head can only be a Sentinel if there is
                  * no element within the list
                  */

                  /* this also means that the list is currently
                   * implicitly fully locked */

                  /* return NULL and let the thread decide what to do */

                  pthread_mutex_unlock(&head_->mutex);

                  return NULL;
            }
            else
            {
                /* the head is not a Sentinel, which implies on of the following:
                 *     - The list has exactly one element, which means:
                 *         => head_ == tail_ && !head_->isSentinel()
                 *     
                 *     OR
                 *     
                 *     - The list has at least two elements, which means:
                 *         => head_ != tail_ && !head_->isSentinel()
                 *
                 *     - The absence of a Sentinel guarantees that the list
                 *       is NOT empty
                 */

                 if(head_ == tail_)
                 {
                      /* single element within the list 
                       * the list is still fully locked, 
                       * implicit because head_ == tail_
                       */

                       /* we replace the only element
                        * with a Sentinel, because
                        * the list is empty afterwards
                        */

                        Node<T>* p = head_;
                        head_ = tail_ = new SentinelNode<T>(NULL, NULL, NULL);

                        pthread_mutex_unlock(&p->mutex);

                        return p;
                 }
                 else
                 {
                      /* at least two elements are in the list
                       * which means that the current head
                       * must have a valid predecessor 
                       */

                       /* Producer and Consumer could
                        * hold a lock on two adjacent
                        * nodes. Thus, the Consumer
                        * must acquire head->prev
                        */

                        Node<T>* p = head_;
                        Node<T>* n = head_->getPrevious();

                        pthread_mutex_lock(&n->mutex);

                        /* head_->prev can now not be owned 
                         * by a producer. 
                         */

                        head_ = n;
                        head_->setNext(NULL);

                        pthread_mutex_unlock(&p->mutex);
                        pthread_mutex_unlock(&n->mutex);

                        return p;
                 } 
            }
        }
};

非常感谢您对这些东西的想法。 我现在正在看它一段时间，也可能完全走错了路。 感谢您的帮助。

谢谢塞巴斯蒂安

Answer 1

假设节点之后仍然是头尾，那么消费和生产都锁定头尾节点。 也就是说，线程可能具有不再是头节点或尾节点的锁定节点。 如果在锁定之前存储了一个“旧”指针，那么您至少会知道锁定了哪个节点，但是锁定的节点可能已经被使用，并且某些逻辑可能不再成立。

我认为锁定头和尾指针比锁定实际节点要容易得多，假设您只需要产生和消费方法。 当然，最简单的解决方案是锁定整个数据结构。