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多线程编程之:Linux线程编程

作者: 时间:2013-09-13 来源:网络 收藏


本文引用地址://m.amcfsurvey.com/article/257101.htm

(3)使用实例。

下面的实例是在9.2.1小节示例代码的基础上增加互斥锁功能,实现原本独立与无序的多个线程能够按顺序执行。


/*thread_mutex.c*/

#include stdio.h>

#include stdlib.h>

#include pthread.h>


#define THREAD_NUMBER 3 /* 线程数 */

#define REPEAT_NUMBER 3 /* 每个线程的小任务数 */

#define DELAY_TIME_LEVELS 10.0 /*小任务之间的最大时间间隔*/

pthread_mutex_t mutex;


void *thrd_func(void *arg)

{

int thrd_num = (int)arg;

int delay_time = 0, count = 0;

int res;

/* 互斥锁上锁 */

res = pthread_mutex_lock(mutex);

if (res)

{

printf(Thread %d lock failedn, thrd_num);

pthread_exit(NULL);

}

printf(Thread %d is startingn, 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 = %dn,

thrd_num, count, delay_time);

}

printf(Thread %d finishedn, thrd_num);

pthread_exit(NULL);

}


int main(void)

{

pthread_t thread[THREAD_NUMBER];

int no = 0, res;

void * thrd_ret;


srand(time(NULL));

/* 互斥锁初始化 */

pthread_mutex_init(mutex, NULL);

for (no = 0; no THREAD_NUMBER; no++)

{

res =(thread[no], NULL, thrd_func, (void*)no);

if (res != 0)

{

printf(Create thread %d failedn, no);

exit(res);

}

}

printf(Create treads successn 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 joinedn, no);

}

else

{

printf(Thread %d join failedn, no);

}

/* 互斥锁解锁 */

pthread_mutex_unlock(mutex);

}

pthread_mutex_destroy(mutex);

return 0;

}


该实例的运行结果如下所示。这里3个线程之间的运行顺序跟创建线程的顺序相同。



$./thread_mutex

Create treads success

Waiting for threads to finish...

Thread 0 is starting

Thread 0: job 0 delay = 7

Thread 0: job 1 delay = 7

Thread 0: job 2 delay = 6

Thread 0 finished

Thread 0 joined

Thread 1 is starting

Thread 1: job 0 delay = 3

Thread 1: job 1 delay = 5

Thread 1: job 2 delay = 10

Thread 1 finished

Thread 1 joined

Thread 2 is starting

Thread 2: job 0 delay = 6

Thread 2: job 1 delay = 10

Thread 2: job 2 delay = 8

Thread 2 finished

Thread 2 joined

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