这是为什么？怎样可以使他从1号或者2号线程开始执行时也能全部执行完3个线程？

Question

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

#define THREAD_NUMBER 3
#define REPEAT_NUMBER 3
#define DELAY_TIME_LEVELS 10.0
pthread_mutex_t mutex;

void *thrd_func(void *arg)
{
int thrd_num = (int)arg;
int delay_time = 0,count =0;
int res;

res = pthread_mutex_lock(&mutex);
if(res)
{
printf("Thread %d lock failed\n",thrd_num);
pthread_exit(NULL);
}
printf("Thread %d is starting\n",thrd_num);

for(count = 0;count <REPEAT_NUMBER;count++)
{
delay_time = (int)(rand() * DELAY_TIME_LEVELS/(RAND_MAX))+1;
sleep(delay_time);
printf("\tThread %d: job %d delay = %d\n",thrd_num,count,delay_time);
}
printf("Thread %d finished\n",thrd_num);
pthread_exit(NULL);

}

int main()
{
pthread_t thread[THREAD_NUMBER];
int no = 0, res;
void * thrd_ret;

srand(time(NULL));

pthread_mutex_init(&mutex,NULL);
for(no = 0;no<THREAD_NUMBER;no++)
{
res = pthread_create(&thread[no],NULL,thrd_func,(void *)no);
if(res != 0)
{
printf("Create thread %d failed\n",no);
exit(res);
}
}
printf("Creat threads success\n Waiting for threads to finish...\n");
for(no=0;no<THREAD_NUMBER;no++)
{
res = pthread_join(thread[no],&thrd_ret);
if(!res)
{
printf("Thread %d joined\n",no);
}
else
{
printf("Thread %d join failed\n",no);
}

pthread_mutex_unlock(&mutex);
}
pthread_mutex_destroy(&mutex);
return 0;
}
这个程序如果从0号线程开始执行的话就会将3个线程全都执行完，如果从1号或2号线程开始执行就会只执行一个线程之后就一直卡在那不动。

Answer 1

这段代码很容易死锁如果1号线程先跑，他lock住mutex，这样其他线程是挂起的，但主线程在等着join 0号线程（顺序执行，先join0,再join1，如此下去），因为0线程此时为挂起，这样主线程也会挂起，而不可能跑去解锁mutex，发生死锁。解决方法是把pthread_mutex_unlock放到线程里面去，线程创建时lock，线程退出时unlock