Pthread:从4个线程中唤醒一个线程

Question

我在C++中唤醒线程时遇到问题。我有4个正在运行的线程。我想在4个正在运行的线程完成后唤醒我的睡眠线程。我做了条件等待手术，但看起来不太好。如何以更高质量的方式完成此过程？

4个任务由广播触发，并同时开始在不同的核心上工作。在每个任务结束时，它将自己的taskID的标志设置为1，并向睡眠任务发送信号。处于睡眠状态的任务每次收到信号时都会唤醒，并检查每个任务的标志。如果4任务标志为1，它将继续并执行自己的工作。

void *thread_sleep( void *arg )
{
pthread_mutex_lock(&mutex_sleep);

while(flag_task[0] == 0 || flag_task[1] == 0 || flag_task[2] == 0 || flag_task[3] == 0)
pthread_cond_wait(&cond_sleep, &mutex_sleep);

/*
.
.
.
.   
*/

flag_task[0] = 0;
flag_task[1] = 0;
flag_task[2] = 0;
flag_task[3] = 0;

pthread_mutex_unlock(&mutex_sleep);
}

void *thread( void *arg)
{
int taskID = *(char *)arg - '0';
while(1)
{
pthread_mutex_lock(&mutex[taskID]);
pthread_cond_wait(&cond, &mutex[taskID]);
/*
.
.
.
.
*/
pthread_mutex_unlock(&mutex[taskID]);
flag_task[taskID] = 1;
pthread_cond_signal(&cond_sleep);
}
}
int main()
{
pthread_create( &pthread1, NULL, thread, (void *)"0" );
pthread_create( &pthread2, NULL, thread, (void *)"1" );
pthread_create( &pthread3, NULL, thread, (void *)"2" );
pthread_create( &pthread4, NULL, thread, (void *)"3" );
pthread_create( &pthread5, NULL, thread_sleep, (void *)"4" );

pthread_cond_broadcast(&cond);
}

Answer 1

我使用屏障解决了问题。谢谢你@Quimby。

void *thread_sleep( void *arg )
{
pthread_mutex_lock(&mutex_sleep);
pthread_cond_wait(&cond_sleep, &mutex_sleep);
/*
.
.
.
.   
*/ 
pthread_mutex_unlock(&mutex_sleep);
}

void *thread( void *arg)
{
int taskID = *(char *)arg - '0';
while(1)
{
pthread_mutex_lock(&mutex[taskID]);
pthread_cond_wait(&cond, &mutex[taskID]);
/*
.
.
.
.
*/
pthread_mutex_unlock(&mutex[taskID]);
pthread_barrier_wait(&barrier);
pthread_cond_signal(&cond_sleep);
}
}
int main()
{
pthread_barrier_init(&barrier, NULL, 4);

pthread_create( &pthread1, NULL, thread, (void *)"0" );
pthread_create( &pthread2, NULL, thread, (void *)"1" );
pthread_create( &pthread3, NULL, thread, (void *)"2" );
pthread_create( &pthread4, NULL, thread, (void *)"3" );
pthread_create( &pthread5, NULL, thread_sleep, (void *)"4" );

sleep(1);
pthread_cond_broadcast(&cond);
}