互斥鎖無數次

Question

據我所知，互斥鎖應該鎖定一次，然後阻止其他人，直到釋放，就像這樣。

但是我的代碼，這似乎是多線程鎖定相同的互斥。我有一個10線程池，所以肯定9應該阻止，1應該鎖定。但我得到這個輸出。

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 

我的互斥體在* .c文件的頂部全局定義，

pthread_mutex_t queuemutex = PTHREAD_MUTEX_INITIALIZER; 

這裏是相關的代碼段。

//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; 
} 

這裏是所有10個線程似乎鎖定的代碼段。

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); 

Answer 1

我的精神力量表明，currentWaiting->status最初0。

由於這種情況，您的代碼將進入while循環並等待條件變量。

等待條件變量解鎖互斥鎖，直到等待完成，允許其他線程獲取它。