c - 用互斥锁重新创建 sem_wait()？

Question

有什么办法可以让我在同一个互斥锁中拥有最多 10 个线程？

类似于 sem_wait() 的值 10。

编辑:

找到这个:

它是信号量的一个实现，使用了互斥量和条件变量。

typedef struct {
  int value, wakeups;
  Mutex *mutex;
  Cond *cond;
} Semaphore;

// SEMAPHORE

Semaphore *make_semaphore (int value)
{
  Semaphore *semaphore = check_malloc (sizeof(Semaphore));
  semaphore->value = value;
  semaphore->wakeups = 0;
  semaphore->mutex = make_mutex ();
  semaphore->cond = make_cond ();
  return semaphore;
}

void sem_wait (Semaphore *semaphore)
{
  mutex_lock (semaphore->mutex);
  semaphore->value--;

  if (semaphore->value < 0) {
    do {
      cond_wait (semaphore->cond, semaphore->mutex);
    } while (semaphore->wakeups < 1);
    semaphore->wakeups--;
  }
  mutex_unlock (semaphore->mutex);
}

void sem_signal (Semaphore *semaphore)
{
  mutex_lock (semaphore->mutex);
  semaphore->value++;

  if (semaphore->value <= 0) {
    semaphore->wakeups++;
    cond_signal (semaphore->cond);
  }
  mutex_unlock (semaphore->mutex);
}

Answer 1

看看是否有帮助

From Book Begining Linux programming a counting semaphore that takes a wider range of values. Normally,semaphores are used to protect a piece of code so that only one thread of execution can run it at any one time. For this job a binary semaphore is needed. Occasionally, you want to permit a limited number of threads to execute a given piece of code; for this you would use a counting semaphore