开发者

How can you implement a condition variable using semaphores?

A while back I was thinking about how to implement various synchronization primitives in terms of one another. For example, in pthrea开发者_开发知识库ds you get mutexes and condition variables, and from these can build semaphores.

In the Windows API (or at least, older versions of the Windows API) there are mutexes and semaphores, but no condition variables. I think that it should be possible to build condition variables out of mutexes and semaphores, but for the life of me I just can't think of a way to do so.

Does anyone know of a good construction for doing this?


Here's a paper from Microsoft Research [pdf] which deals with exactly that.


One way of implementing X given semaphores is to add a server process to the system, use semaphores to communicate with it, and have the process do all the hard work of implementing X. As an academic exercise, this might be cheating, but it does get the job done, and it can be more robust to misbehaviour by the client processes, or to their sudden death.


I may be missing something here, but there seem to be a simpler way to implement a Condition from a Semaphore and Lock than the way described in the paper.

class Condition {
    sem_t m_sem;
    int   m_waiters;
    int   m_signals;
    pthread_mutex_t *m_mutex;
public:

    Condition(pthread_mutex_t *_mutex){
        sem_init(&this->m_sem,0,0);
        this->m_waiters = 0;
        this->m_signals = 0;
        this->m_mutex = _mutex;
    }
    ~Condition(){}
    void wait();
    void signal();
    void broadcast();
};

void Condition::wait() {
    this->m_waiters++;
    pthread_mutex_unlock(this->m_mutex);
    sem_wait(&this->m_sem);
    pthread_mutex_lock(this->m_mutex);
    this->m_waiters--;
    this->m_signals--;
}

void Condition::signal() {
    pthread_mutex_lock(this->m_mutex);
    if (this->m_waiters && (this->m_waiters > this->m_signals)) {
        sem_post(&this->m_sem);
        this->m_signals++;
    }
    pthread_mutex_unlock(this->m_mutex);
}
0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜