C 读写器线程锁解锁

Question

我试图完成我认为很简单的任务，但几天后已经达到了临界点。

我正在尝试使用多个读取器和单个写入器来模拟数据库。当我运行程序时它死锁。

我试图将其基于此算法:

 1 semaphore accessMutex;     // Initialized to 1
 2 semaphore readersMutex;    // Initialized to 1
 3 semaphore orderMutex;      // Initialized to 1
 4 
 5 unsigned int readers = 0;  // Number of readers accessing the resource
 6 
 7 void reader()
 8 {
 9   P(orderMutex);           // Remember our order of arrival
10 
11   P(readersMutex);         // We will manipulate the readers counter
12   if (readers == 0)        // If there are currently no readers (we came first)...
13     P(accessMutex);        // ...requests exclusive access to the resource for readers
14   readers++;               // Note that there is now one more reader
15   V(orderMutex);           // Release order of arrival semaphore (we have been served)
16   V(readersMutex);         // We are done accessing the number of readers for now
17 
18   ReadResource();          // Here the reader can read the resource at will
19 
20   P(readersMutex);         // We will manipulate the readers counter
21   readers--;               // We are leaving, there is one less reader
22   if (readers == 0)        // If there are no more readers currently reading...
23     V(accessMutex);        // ...release exclusive access to the resource
24   V(readersMutex);         // We are done accessing the number of readers for now
25 }
26 
27 void writer()
28 {
29   P(orderMutex);           // Remember our order of arrival
30   P(accessMutex);          // Request exclusive access to the resource
31   V(orderMutex);           // Release order of arrival semaphore (we have been served)
32 
33   WriteResource();         // Here the writer can modify the resource at will
34 
35   V(accessMutex);          // Release exclusive access to the resource
36 }

但是我尝试仅使用 pthreads 而不是信号量来实现它。正如你所预料的，结果是这样的:

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#define MAX 5 //size of buffer

pthread_mutex_t mutex; //pthread_mutex type
pthread_cond_t condw, condr; //reader/writer cond var
int buffer = 0, rc = 0;

pthread_mutex_t db; //pthread_mutex type

//reader-writer header:
void* writer(void* ptr);
void* reader(void* ptr);

void use_data();
void write_database();
void readdb();

int main(int argc, char** argv)
{
  pthread_t read,write;
  //allow signal back and forth
  pthread_mutex_init(&mutex,0); //init mutex
  pthread_cond_init(&condw, 0); //init writer
  pthread_cond_init(&condr,0); //init reader

    pthread_mutex_init(&db,0); //init db

  //this calls the void* reader function
  pthread_create(&write,0,writer,0); //create thread
  pthread_create(&read,0,reader,0); //create thread

  //let them join
  pthread_join(read,0);
  pthread_join(write,0);

  //destroy them
  pthread_cond_destroy(&condw);
  pthread_cond_destroy(&condr);
  pthread_mutex_destroy(&db);
  pthread_mutex_destroy(&mutex);

  return 0;
}//end main

void* reader(void* arg)
{
  while(1) //if there is a reader lock the db 
  {
    pthread_mutex_lock(&mutex);
    rc++;
    if(rc==1)
    {
      pthread_mutex_lock(& db);
    }
    pthread_mutex_unlock(&mutex);
    readdb();
    pthread_mutex_lock(&mutex);
    rc--;
    if(rc==0) 
    {
      pthread_mutex_unlock(&db);
    }
    pthread_mutex_unlock(&mutex);
    use_data();
  }
}

void* writer(void* arg)
{
  while(1) //unlock the db
  {
    pthread_mutex_lock(&db);
    write_database();
    pthread_mutex_unlock(&db);
  }
}

void use_data(){}
void write_database(){}
void readdb(){}

任何有关工作解决方案和解释我们出错的地方的帮助都将受到赞赏，因为这将帮助我和我的同事摆脱困境。问候。

Answer 1

问题在于您的源算法依赖于这样一个事实:一个线程锁定了 accessMutex 锁，然后另一个线程可能会解锁它。这对于基于信号量的互斥体是允许的，但对于 pthreads 互斥体是不允许的。

pthread 中有信号量，由 sem_init()、sem_post() 和 sem_wait() 函数提供。您可以使用它们来编写源算法的直接实现，并且它应该可以正常工作。

另外，pthreads 还提供 native 读写锁类型 - 请参阅函数 pthread_rwlock_init()、pthread_rwlock_rdlock()、pthread_rwlock_wrlock() 和 pthread_rwlock_unlock()。您可以使用它来进行非常简单的实现，但如果它应该是一个学习练习，那么显然这是没有意义的。