boost :: threads notify_one vs notify_all的性能

Question

我有一个Semaphore实现来使用boost :: threads来管理共享资源。我对信号量的实现如下所示。

void releaseResource()
{
    boost::unique_lock<boost::mutex> lock(mutex_);
    boost::thread::id curr_thread =  boost::this_thread::get_id();
    // scan through to identify the current thread
    for (unsigned int i=0; i<org_count;++i)
    {
        if (res_data[i].thread_id == curr_thread)
        {
            res_data[i].thread_id = boost::thread::id();
            res_data[i].available = true;
            break;
        }
    }
    ++count_;
    condition_.notify_one();
}

unsigned int acquireResource()
{
    boost::unique_lock<boost::mutex> lock(mutex_);
    while(count_ == 0)  { // put thread to sleep until resource becomes available
        condition_.wait(lock);
    }
    --count_;
    // Scan through the resource list and hand a index for the resource
    unsigned int res_ctr;
    for (unsigned int i=0; i<org_count;++i)
    {
        if (res_data[i].available)
        {
            res_data[i].thread_id = boost::this_thread::get_id();
            res_data[i].available = false;
            res_ctr = i;    
            break;
        }
    }
    return res_ctr;
}

我的问题是当线程数多于可用资源数时，我注意到性能下降。如果我使用notify_all来唤醒线程而不是notify_one，如releaseResource()代码中所示，我看到性能提高了。还有其他人经历过类似的事吗？

我正在使用 Windows 7 和提升1.52 。