c - 多次互斥锁定

Question

据我所知，互斥量应该锁定一次然后阻塞其他直到被释放，就像这样。

但是在我的代码中，似乎多个线程正在锁定同一个互斥量。我有一个 10 线程池，所以 9 个应该阻塞，1 个应该锁定。但是我得到了这个输出。

Thread 0 got locked 
Thread 1 got locked 
Thread 3 got locked 
Thread 4 got locked 
Thread 2 got locked 
Thread 5 got locked 
Thread 6 got locked 
Thread 7 got locked 
Thread 8 got locked 
Thread 9 got locked 

我的互斥量在 *.c 文件的顶部全局定义为，

pthread_mutex_t queuemutex = PTHREAD_MUTEX_INITIALIZER;

这里是相关的代码段。

    //In the main function which creates all the threads
int k;
for (k = 0; k < POOLSIZE; k++) {
    pthread_t thread;
    threadinformation *currentThread = (threadinformation *)malloc(sizeof(threadinformation));
    currentThread->state = (int *)malloc(sizeof(int));
    currentThread->state[0] = 0;
    currentThread->currentWaiting = currentWaiting;
    currentThread->number = k;
    threadArray[k] = currentThread;
    pthread_create(&thread, NULL, readWriteToClient, threadArray[k]);
    currentThread->thread = thread;
    joinArray[k] = thread;
}

这是一段代码，其中所有 10 个线程似乎都获得了锁。

pthread_mutex_lock(&queuemutex);

fprintf(stderr,"Thread %d got locked \n",threadInput->number);

while((threadInput->currentWaiting->status) == 0){
    pthread_cond_wait(&cond, &queuemutex);
    fprintf(stderr,"Thread %d got signalled \n",threadInput->number);
}

connfd = threadInput->currentWaiting->fd;
threadInput->currentWaiting->status = 0;
pthread_cond_signal(&conncond);
pthread_mutex_unlock(&queuemutex);

Answer 1

我的灵能表明 currentWaiting->status 最初是 0。

由于是这种情况，您的代码进入 while 循环并等待条件变量。

等待条件变量解锁互斥量直到等待完成，允许其他线程获取它。