【发布时间】：2018-06-04 22:41:56
【问题描述】：
我真的很难理解如何以让线程同时运行的方式锁定和解锁互斥锁。现在我试图让每个线程都做一些事情，然后等到所有线程都准备好，然后重复它。这应该反复发生，所以我把它全部放在一个条件循环中。我的问题是，似乎我的广播从未被任何线程接收，因为等待信号的每个线程都会永远等待，而发送信号的线程会继续。
我已经简化了我的代码，使其尽可能简单，同时仍然可以编译和运行：
#include <stdio.h>
#include <pthread.h>
#include <unistd.h>
#define num_threads 4
int repeat = 1;
int timesLooped = 0;
int waitingThreads = 0;
int threadNo = -1;
pthread_mutex_t mutex1;
pthread_cond_t condition;
//thread body
void* threadBody(void* arg){
//sensitive info: requires lock
if(pthread_mutex_lock(&mutex1)){ printf("error1\n"); }
threadNo++;
int threadId = threadNo;
//other info is set here as well in my code
if(pthread_mutex_unlock(&mutex1)){ printf("error2\n"); }
//main loop in the thread body
while(repeat == 1){
if(pthread_mutex_lock(&mutex1)){ printf("error3\n"); }
//wait until all threads are ready
while(waitingThreads < num_threads - 1){
printf(" %d! ", threadId);
waitingThreads++;
pthread_cond_wait(&condition, &mutex1);
printf(" %d ", threadId);
if(pthread_mutex_unlock(&mutex1)){ printf("error4\n"); }
}
//last thread will broadcast
if(waitingThreads == num_threads - 1){
printf("\n\nthread %d was last! broadcasting to let everyone proceed...", threadId);
timesLooped++;
waitingThreads = 0;
if(timesLooped == 3){
repeat = 0;
}
sleep(1);
pthread_cond_broadcast(&condition);
if(pthread_mutex_unlock(&mutex1)){ printf("error5\n"); }
}
}
printf("\n\nexiting thread %d\n", threadId);
pthread_exit((void*) arg);
}
int main(int argc, char** argv){
pthread_t threads[num_threads];
void* retval;
//create threads
for(long i = 0; i < num_threads; i++){
pthread_create(&threads[i], NULL, threadBody, (void*) i);
}
//join threads
for(long j = 0; j < num_threads; j++){
pthread_join(threads[j], &retval);
}
printf("\n\nDONE\n");
}
这给了我一个输出：
线程 3 是最后一个！广播让大家继续...
线程 2 是最后一个！广播让大家继续...
线程 1 是最后一个！广播让大家继续...
退出线程1（死锁，其他线程永不退出）
【问题讨论】：
-
如果您提供实际的可编译测试（您提供的代码无效且无法编译），您会得到更好的答案。不加倍间距的奖励积分。
-
我已将代码更新为可编译。