C++ pthread cond_wait 和 cond_broadcast的使用

转自:https://blog.csdn.net/wangzhiyu1980/article/details/44227913

参考:

https://blog.csdn.net/YEYUANGEN/article/details/37593533

https://www.cnblogs.com/secondtonone1/p/5580203.html

一个简单的实例程序,说明pthread_cond_wait 和 pthread_cond_broadcast 的使用方式。

函数定义:

int pthread_cond_wait(pthread_cond_t *cond, pthread_mutex_t *mutex)

int pthread_cond_broadcast(pthread_cond_t *cond)

函数说明:

pthread_cond_wait :有两个输入参数,一个是pthread_cond_t,是函数将要等待的信号,另一个是 pthread_mutex_t,一个互斥锁。用于对信号量进行保护,防止多个线程同时对其进行操作。在线程开始等待信号量前,必须由本线程对互斥锁进行锁定,然后pthread_cond_wait会更新条件等待队列,并且释放互斥量,允许其他线程进行访问;当cond 满足条件允许线程继续执行时,wait_cond也会先对mutex 进行锁定,对cond进行处理,然后再允许线程继续运行。所以pthread_cond_wait() 后的pthread_mutex_unlock()还是必要的。

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实例程序:

#include <pthread.h>

#include <iostream>

#include <unistd.h>



using namespace std;


static pthread_mutex_t mtx=PTHREAD_MUTEX_INITIALIZER;

static pthread_cond_t cond=PTHREAD_COND_INITIALIZER;


static void* func_1(void* arg)

{

cout << "func_1 start" << endl;


pthread_mutex_lock(&mtx);

cout << "func_1 lock mtx" << endl;


cout << "func_1 wait cond" << endl;

pthread_cond_wait(&cond, &mtx);


cout << "func_1 unlock mtx" << endl;

pthread_mutex_unlock(&mtx);


cout << "func_1 end" << endl;

sleep(5);


return NULL;

}


static void* func_2(void* arg)

{

cout << "func_2 start" << endl;


pthread_mutex_lock(&mtx);

cout << "func_2 lock mtx" << endl;


cout << "func_2 wait cond" << endl;

pthread_cond_wait(&cond, &mtx);


cout << "func_2 unlock mtx" << endl;

pthread_mutex_unlock(&mtx);


cout << "func_2 end" << endl;

sleep(5);


return NULL;

}



int main()

{

pthread_t tid1, tid2;


cout << "main create thread" << endl;

pthread_create(&tid1, NULL, func_1, NULL);

pthread_create(&tid2, NULL, func_2, NULL);


sleep(3);

cout << "main boradcast signal" << endl;

pthread_cond_broadcast(&cond);


cout << "main join thread" << endl;


pthread_join(tid1, NULL);

pthread_join(tid2, NULL);


cout << "main end" << endl;

return 0;

}


 

测试结果:

pic_1

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转载自blog.csdn.net/fchyang/article/details/81202877