我正在玩一个关于MPMC FIFO-Queue的双向链表（主要用于演示目的）。我的目标是现在试图将我的错误记录一段时间，但我并没有真正取得进展。
在所有Producer-Threads生成所有值之后不久，我遇到了死锁。我将问题归结为以下内容：
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addr 0x7fb380c039d0等待mutx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"
}
不幸的是我不能昆虫pthread_mutex_t，因为它是一种不透明的类型。 （虽然我在GDB中打开了不透明类型？）。否则，很高兴看到该互斥锁的所有者目前是谁。
我的清单代码如下 - 它被过度评论，以解释我对实施的看法。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。我现在正在看它一段时间，也可能完全走错了路。感谢帮助。
谢谢， 塞巴斯蒂安
答案 0 :(得分：0)
消耗和生成都锁定头部和尾部节点，假设节点之后仍然是头部和尾部。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，您至少会知道锁定了哪个节点，但锁定的节点可能已经被占用，并且您的某些逻辑可能不再存在。
我认为，假设您只需要生成和使用方法，锁定头部和尾部指针而不是实际节点会更容易。当然，最简单的解决方案是锁定整个数据结构。