c++ - 并发双向链表 - 多个生产者/消费者 FIFO 队列 - 死锁

Question

我正在研究 MPMC FIFO 队列的双向链表(主要用于演示目的)。我现在的目标是暂时纠正我的错误，但我并没有真正取得进展。

在所有 Producer-Threads 产生所有值后不久，我遇到了死锁。我将问题归结为以下几点:

锁定获取:

<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，并且在地址 0x7fb380c039d0 等待互斥锁的唯一线程是线程 2(唯一的消费者线程):

    (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"
}

不幸的是我不能插入 pthread_mutex_t 因为它是一个不透明的类型。 (尽管我在 GDB 中打开了不透明类型？)。否则，最好看看该互斥锁当前的所有者是谁。

我的列表代码如下 - 注释过多，以解释我对实现的想法。请注意，到目前为止，这不是一个好的或高性能的实现 - 它主要用于说明目的。

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;
                 } 
            }
        }
};

我真的很感激你对这些东西的想法和想法。我正在研究它一段时间，也可能完全走错了路。感谢帮助..

谢谢，塞巴斯蒂安

Answer 1

consume 和 produce 都锁定头节点和尾节点，假设节点之后仍然是头节点和尾节点。也就是说，线程可能已锁定不再是头节点或尾节点的节点。如果您在锁定之前存储了一个“旧”指针，您至少会知道您锁定了哪个节点，但锁定的节点可能已经被消耗，并且您的某些逻辑可能不再成立。

我认为锁定头指针和尾指针比锁定实际节点更容易，假设您只需要生产和消费方法。当然，最简单的解决方案是锁定整个数据结构。