pthread 中的唤醒丢失

Question

我写了一些程序来尝试 pthread 条件等待。 但问题是不能保证发出信号时会被捕获，从而导致线程失去唤醒。我该如何解决这个问题？

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

pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;

void *thread_func1(void* arg){
  printf("thread1 started\n");
  pthread_mutex_lock(&mutex);
  printf("thread1: signalling\n");
  pthread_cond_signal(&cond);
  printf("thread1: signalled\n");
  pthread_mutex_unlock(&mutex);
  printf("thread1: exiting\n");
  pthread_exit(0);
}

void *thread_func2(void* arg){
  printf("thread2 started\n");
  pthread_mutex_lock(&mutex);
  printf("thread2: waiting for signal..\n");
  pthread_cond_wait(&cond, &mutex);
  printf("thread2: signal received\n");
  pthread_mutex_unlock(&mutex);
  printf("thread2: exiting\n");
  pthread_exit(0);
}

int main(int argc, char** argv){
  pthread_t thread1, thread2;

  pthread_create(&thread1, NULL, thread_func1, NULL);
  pthread_create(&thread2, NULL, thread_func2, NULL);

  pthread_join(thread1, 0);
  pthread_join(thread2, 0);

  return 0;
}

这是运行的输出：

thread1 started
thread1: signalling
thread2 started
thread2: waiting for signal..
thread1: signalled
thread1: exiting
// nothing happens now; where is the signal??

这是另一个（有效）：

thread2 started
thread2: waiting for signal..
thread1 started
thread1: signalling
thread1: signalled
thread1: exiting
thread2: signal received
thread2: exiting
// program successfully exits

我现在不关心任何类型的临界区，所以我没有使用任何锁。

我如何确保这个东西每次运行都有效？

编辑：我已经按照下面 alk 的回答编辑了代码。我添加了初始化程序和锁。我发布的原始代码是here。

Answer 1

正如您所注意到的，线程 1 可能会在线程 2 调用 pthread_cond_wait() 之前发出条件变量的信号。条件变量不会“记住”它已发出信号，因此唤醒将丢失。因此，您需要使用某种变量来确定线程 2 是否需要等待。

int signalled = 0;

void *thread_func1(void* arg){
  printf("thread1 started\n");
  pthread_mutex_lock(&mutex);
  printf("thread1: signalling\n");
  signalled = 1;
  pthread_cond_signal(&cond);
  printf("thread1: signalled\n");
  pthread_mutex_unlock(&mutex);
  printf("thread1: exiting\n");
  pthread_exit(0);
}

void *thread_func2(void* arg){
  printf("thread2 started\n");
  pthread_mutex_lock(&mutex);
  printf("thread2: waiting for signal..\n");
  if(!signalled) {
    pthread_cond_wait(&cond, &mutex);
  }
  printf("thread2: signal received\n");
  pthread_mutex_unlock(&mutex);
  printf("thread2: exiting\n");
  pthread_exit(0);
}

但是，此代码仍然不正确。 pthreads 规范指出条件变量上可能会发生“虚假唤醒”。这意味着即使没有人调用pthread_cond_signal() 或pthread_cond_broadcast()，pthread_cond_wait() 也可能返回。因此，您需要在循环中检查标志，而不是只检查一次：

void *thread_func2(void* arg){
  printf("thread2 started\n");
  pthread_mutex_lock(&mutex);
  printf("thread2: waiting for signal..\n");
  while(!signalled) {
    pthread_cond_wait(&cond, &mutex);
  }
  printf("thread2: signal received\n");
  pthread_mutex_unlock(&mutex);
  printf("thread2: exiting\n");
  pthread_exit(0);
}

更新：结合条件变量和计数器功能的另一种方法是使用信号量。

pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
sem_t sem;

void *thread_func1(void* arg){
  printf("thread1 started\n");
  printf("thread1: signalling\n");
  sem_post(&sem);
  printf("thread1: signalled\n");
  printf("thread1: exiting\n");
  pthread_exit(0);
}

void *thread_func2(void* arg){
  printf("thread2 started\n");
  printf("thread2: waiting for signal..\n");
  sem_wait(&sem);
  printf("thread2: signal received\n");
  printf("thread2: exiting\n");
  pthread_exit(0);
}

int main(int argc, char** argv){
  pthread_t thread1, thread2;

  sem_init(&sem);

  pthread_create(&thread1, NULL, thread_func1, NULL);
  pthread_create(&thread2, NULL, thread_func2, NULL);

  pthread_join(thread1, 0);
  pthread_join(thread2, 0);

  sem_destroy(&sem);

  return 0;
}

Answer 2

对于初学者，您的代码未初始化mutex 和cond。

要这样做，至少要这样做：

pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;

---

传递给pthread_cond_wait() 的互斥锁也应锁定。当函数返回时，它被锁定。

---

更新：

您的代码引入了比赛。也就是说：如果线程 1 在线程 2 等待信号之前发出信号，线程 2 将永远等待，这很可能发生在您显示的第一个跟踪中。