互斥锁的适当粒度

Question

我正在开发一个程序，该程序打印来自搜索文件系统的多个线程的regex匹配，每个线程将其id添加到队列中，并通知主线程有一些线程要从线程池（std：：vector）中删除。

void RegexSearch::Run()
{
    running = true;
    search_pool.push_back(std::thread([&] { StartPrint(); }));

    /*
    Get directories and create another thread for each subdirectory while there are
    less threads than maximum in the pool.
    */

    fs::path searchDir(directory);

    for (const auto& dir : searchDir)
    {
        if (search_pool.size() >= MAX_THREADS)
        {
            std::unique_lock<std::mutex> lck(mu, std::defer_lock);
            // Wait to be notified that thread is finished execution and another can start.
            max_thread_cond.wait(lck);

            lck.lock();
            while (!finished_threads.empty())
            {
                lck.unlock();

                lck.lock();
                std::thread::id remove_thread = finished_threads.front();
                lck.unlock();

                lck.lock();
                finished_threads.pop();
                lck.unlock();

                std::remove_if(search_pool.begin(), search_pool.end(), [&remove_thread](const std::thread::id &t) { return remove_thread == t; });
                lck.lock();
            }
        }
    }
}

Answer 1

您当然必须保留！empty()的测试锁，直到您从队列中删除了该项。否则，几个线程可能会尝试移除同一个项--也许是最后一个。

我还看到函数开始处的push_back不受保护。如果可能有多个线程访问search_pool，则必须对其进行保护。