Question

当我查看主题时，我找到了来自here的代码。如您所见，两个函数线程使用相同的mutex。那么，即使先前的线程拥有互斥锁，如何发信号通知其他线程或捕获信号并继续其功能？怎么/为什么没有死锁？这有点令人困惑。

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>

pthread_mutex_t count_mutex     = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t  condition_var   = PTHREAD_COND_INITIALIZER;

void *functionCount1();
void *functionCount2();
int  count = 0;
#define COUNT_DONE  10
#define COUNT_HALT1  3
#define COUNT_HALT2  6

main()
{
   pthread_t thread1, thread2;

   pthread_create( &thread1, NULL, &functionCount1, NULL);
   pthread_create( &thread2, NULL, &functionCount2, NULL);

   pthread_join( thread1, NULL);
   pthread_join( thread2, NULL);

   printf("Final count: %d\n",count);

   exit(EXIT_SUCCESS);
}

// Write numbers 1-3 and 8-10 as permitted by functionCount2()

void *functionCount1()
{
   for(;;)
   {
      // Lock mutex and then wait for signal to relase mutex
      pthread_mutex_lock( &count_mutex );     //     <---- Same mutex

      // Wait while functionCount2() operates on count
      // mutex unlocked if condition varialbe in functionCount2() signaled.
      pthread_cond_wait( &condition_var, &count_mutex );
      count++;
      printf("Counter value functionCount1: %d\n",count);

      pthread_mutex_unlock( &count_mutex );

      if(count >= COUNT_DONE) return(NULL);
    }
}

// Write numbers 4-7

void *functionCount2()
{
    for(;;)
    {
       pthread_mutex_lock( &count_mutex );    //     <---- Same mutex

       if( count < COUNT_HALT1 || count > COUNT_HALT2 )
       {
          // Condition of if statement has been met. 
          // Signal to free waiting thread by freeing the mutex.
          // Note: functionCount1() is now permitted to modify "count".
          pthread_cond_signal( &condition_var );
       }
       else
       {
          count++;
          printf("Counter value functionCount2: %d\n",count);
       }

       pthread_mutex_unlock( &count_mutex );

       if(count >= COUNT_DONE) return(NULL);
    }

}

Answer 1

如果一个主题拥有锁定，然后另一个人在另一个人释放它之前无法获得它。
现在，如果线程A首先获得锁定，它将阻止线程B 在关键部分进展，直到锁定被释放。

为什么没有死锁？

阅读this：

firstname = obj.employees[1][fn]  
lastname = obj.employees[1][ln]

此函数以原子方式释放互斥锁并导致调用线程阻塞条件变量cond;

线程A会在命中int pthread_cond_wait(pthread_cond_t *cond, pthread_mutex_t *mutex);时释放锁定，线程B可以进行并发出条件变量信号。

Pthread条件变量，没有死锁

1 个答案: