Question

我是pthreads的新手，我编写了一个示例程序来利用pthread中的条件事件。

Thread2只是没有退出......我猜的代码意图非常明确。

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

pthread_mutex_t mut;
pthread_cond_t con;

#define  LOCK &mut
#define  COND &con

#define HAPPY 1
#define SAD 0

int count = 0, response = SAD;

void func_t1();
void func_t2();

int main()
{
  pthread_t tid1, tid2;

  pthread_create(&tid1, NULL, &func_t1, NULL );
  pthread_create(&tid2, NULL, &func_t2, NULL );

  pthread_join(tid1, NULL );
  pthread_join(tid2, NULL );

  return 0;
}

void func_t1()
{
  for (;;)
  {
    pthread_mutex_lock(LOCK);

    pthread_cond_wait(COND, LOCK);
    if (count == 10)
    {
      printf("Thread1: You did your job finally, get lost now");
      response = HAPPY;
      pthread_mutex_unlock(LOCK);
      break;
    }
    else
    {
      printf("Thread1: You are not capable of making me happy");

      pthread_mutex_unlock(LOCK);
    }
  }
}

void func_t2()
{
  for (;;)
  {
    if (response == SAD)
    {
      pthread_mutex_lock(LOCK);
      count++;
      printf("thread2: count incremented to %d...pls check if you are happy \n",
          count);
      pthread_cond_signal(COND);
      pthread_mutex_unlock(LOCK);
    }
    else
    {
      printf("thread2:Ha..I finally made her happy \n");
      break;
    }
  }
}

输出：

thread2: count incremented to 106927...pls check if you are happy 
thread2: count incremented to 106928...pls check if you are happy 
thread2: count incremented to 106929...pls check if you are happy 
thread2: count incremented to 106930...pls check if you are happy 
thread2: count incremented to 106931...pls check if you are happy 
thread2: count incremented to 106932...pls check if you are happy 
thread2: count incremented to 106933...pls check if you are happy 
thread2: count incremented to 106934...pls check if you are happy 
thread2: count incremented to 106935...pls check if you are happy 
thread2: count incremented to 106936...pls check if you are happy 
thread2: count incremented to 106937...pls check if you are happy 
thread2: count incremented to 106938...pls check if you are happy 
thread2: count incremented to 106939...pls check if you are happy 
thread2: count incremented to 106940...pls check if you are happy 
thread2: count incremented to 106941...pls check if you are happy 
thread2: count incremented to 106942...pls check if you are happy 
thread2: count incremented to 106943...pls check if you are happy 
Thread1: You are not capable of making me happythread2: count incremented to 106944...pls check if you are happy 
thread2: count incremented to 106945...pls check if you are happy 
thread2: count incremented to 106946...pls check if you are happy 
thread2: count incremented to 106947...pls check if you are happy 
thread2: count incremented to 106948...pls check if you are happy 
thread2: count incremented to 106949...pls check if you are happy 
thread2: count incremented to 106950...pls check if you are happy 
thread2: count incremented to 106951...pls check if you are happy 
thread2: count incremented to 106952...pls check if you are happy 
thread2: count incremented to 106953...pls check if you are happy 
thread2: count incremented to 106954...pls check if you are happy 
thread2: count incremented to 106955...pls check if you are happy 
thread2: count incremented to 106956...pls check if you are happy 
thread2: count incremented to 106957...pls check if you are happy 
thread2: count incremented to 106958...pls check if you are happy 
thread2: count incremented to 106959...pls check if you are happy

它永远不会结束......

如果您能指出上述代码中的缺陷，将会有所帮助。

Answer 1

所以你基本上有一个包含两个线程的程序：

第一个线程正在等待计数到来10.它会从线程2的信号中唤醒。

线程2递增计数变量并发送信号，然后重复。

由于线程2不等待线程1的回复，因此没有任何东西可以保证线程1会看到每个更改都要计数。

相反它可能看起来像：

Thread 2 increments count & signals
Thread 2 increments count & signals
Thread 2 increments count & signals
Thread 2 increments count & signals
Thread 1 sees signal - not 10
Thread 2 increments count & signals
Thread 2 increments count & signals
Thread 2 increments count & signals
Thread 1 sees signal - not 10
...
...

如果线程1检查了count的当前值并且是SAD，则需要添加仅允许线程2递增计数的机制。

Pthread同步查询

1 个答案: