据我所知，互斥锁应该锁定一次，然后阻止其他人直到被释放，就像这样。
但是使用我的代码，似乎多个线程正在锁定同一个互斥锁。我有一个 10 个线程池，所以肯定 9 个应该阻塞，1 个应该锁定。但我得到了这个输出。
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
我的互斥锁在 *.c 文件的顶部全局定义为，
pthread_mutex_t queuemutex = PTHREAD_MUTEX_INITIALIZER;
这是相关的代码段。
//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;
}
这是所有 10 个线程似乎都获得锁的代码段。
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);