c - 试图理解 pthread_cond_signal()
问题描述
嗨,我是 C 新手,试图了解互斥锁、条件和线程。我有线程如何工作的基础知识。如果我错了,请纠正我,据我了解,一个线程正在等待另一个 2 个线程发送信号以唤醒他。由于互斥锁,代码一次只能由一个线程运行,线程 2 获取锁并运行代码,sleep(1)
有助于交替线程,因此线程 2 不会再次运行它,因此线程 3 获取锁并运行运行代码等等。只有锁定和释放互斥锁的线程。一旦它到达COUNT_LIMIT 12
,它就会向线程 1 的等待线程发送一个信号。如果我没有误解任何内容,我到此为止
所以我的问题是这几段代码在void *watch_count
:
pthread_mutex_lock(&count_mutex);
pthread_cond_wait(&count_threshold_cv, &count_mutex);
pthread_mutex_unlock(&count_mutex);
它是否获得与 in 相同的锁void *inc_count
?如果是这样,它会锁定互斥体并且pthread_cond_wait
正在等待信号void *watch_count
,然后继续执行代码?pthread_mutex_lock
和pthread_mutex_unlock
在做什么,这是否void *watch_count
在等待线程 1 完成工作时阻止线程 2 和 3 执行?
我希望我能理解或解释我正确理解的内容,提前感谢您提供的任何建议!
#define NUM_THREADS 3
#define TCOUNT 10
#define COUNT_LIMIT 12
int count = 0;
pthread_mutex_t count_mutex;
pthread_cond_t count_threshold_cv;
void *inc_count(void *t)
{
int i;
long my_id = (long)t;
for (i=0; i < TCOUNT; i++) {
pthread_mutex_lock(&count_mutex);
count++;
/*
Check the value of count and signal waiting thread when condition is
reached. Note that this occurs while mutex is locked.
*/
if (count == COUNT_LIMIT) {
printf("inc_count(): thread %ld, count = %d Threshold reached. ",
my_id, count);
pthread_cond_signal(&count_threshold_cv);
printf("Just sent signal.\n");
}
printf("inc_count(): thread %ld, count = %d, unlocking mutex, i: %d\n",
my_id, count, i);
pthread_mutex_unlock(&count_mutex);
/* Do some work so threads can alternate on mutex lock */
sleep(1);
}
pthread_exit(NULL);
}
void *watch_count(void *t)
{
long my_id = (long)t;
printf("Starting watch_count(): thread %ld\n", my_id);
pthread_mutex_lock(&count_mutex);
while (count < COUNT_LIMIT) {
printf("watch_count(): thread %ld Count= %d. Going into wait...\n", my_id,count);
pthread_cond_wait(&count_threshold_cv, &count_mutex);
printf("watch_count(): thread %ld Condition signal received. Count= %d\n", my_id,count);
printf("watch_count(): thread %ld Updating the value of count...\n", my_id,count);
count += 125;
printf("watch_count(): thread %ld count now = %d.\n", my_id, count);
}
printf("watch_count(): thread %ld Unlocking mutex.\n", my_id);
pthread_mutex_unlock(&count_mutex);
pthread_exit(NULL);
}
int main(int argc, char *argv[])
{
int i, rc;
long t1=1, t2=2, t3=3;
pthread_t threads[3];
pthread_attr_t attr;
/* Initialize mutex and condition variable objects */
pthread_mutex_init(&count_mutex, NULL);
pthread_cond_init (&count_threshold_cv, NULL);
/* For portability, explicitly create threads in a joinable state */
pthread_attr_init(&attr);
pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);
pthread_create(&threads[0], &attr, watch_count, (void *)t1);
pthread_create(&threads[1], &attr, inc_count, (void *)t2);
pthread_create(&threads[2], &attr, inc_count, (void *)t3);
/* Wait for all threads to complete */
for (i = 0; i < NUM_THREADS; i++) {
pthread_join(threads[i], NULL);
}
printf ("Main(): Waited and joined with %d threads. Final value of count = %d. Done.\n",
NUM_THREADS, count);
/* Clean up and exit */
pthread_attr_destroy(&attr);
pthread_mutex_destroy(&count_mutex);
pthread_cond_destroy(&count_threshold_cv);
pthread_exit (NULL);
}
解决方案
是的,您所有的线程都在获取相同的锁(该程序中只有一个互斥锁 - count_mutex
)。
watch_count()
调用时pthread_cond_wait()
,它会解锁互斥锁并在一个原子操作中暂停。这允许其他线程获取锁。当被调用的线程pthread_cond_wait()
唤醒时,它会在pthread_cond_wait()
返回之前重新获取锁(这意味着如果另一个线程已经锁定了互斥锁,pthread_cond_wait()
那么直到另一个线程首先释放锁才能返回)。
调用时必须锁定互斥锁pthread_cond_wait()
- 这是 API 的要求。此外,watch_count()
线程锁定count_mutex
以便它可以访问count
变量而不会引入竞争条件(count
来自另一个线程的并发修改)。请注意,没有一个线程在没有锁定的情况下访问该count
变量 - 无论是读取还是写入。count_mutex
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