c++ - 如果没有线程需要唤醒，是否需要获取锁并通知condition_variable？

Question

我正在阅读 this reference并看到:

The thread that intends to modify the variable has to

1. acquire a std::mutex (typically via std::lock_guard)

2. perform the modification while the lock is held

3. execute notify_one or notify_all on the std::condition_variable (the lock does not need to be held for notification)

如果更改不需要唤醒线程，如 on_pause函数在这里，为什么需要获取锁(1)或调用通知(3)？ (只是叫醒他们说晚安？)

std::atomic<bool> pause_;
std::mutex pause_lock_;
std::condition_variable pause_event_;

void on_pause() // part of main thread
{
    // Why acquiring the lock is necessary?
    std::unique_lock<std::mutex> lock{ pause_lock_ };
    pause_ = true;
    // Why notify is necessary?
    pause_event_.notify_all();
}

void on_resume() // part of main thread
{
    std::unique_lock<std::mutex> lock{ pause_lock_ };
    pause = false;
    pause_event_.notify_all();
}

void check_pause() // worker threads call this
{
    std::unique_lock<std::mutex> lock{ pause_lock_ };
    pause_event_.wait(lock, [&](){ return !pause_; });
}

Answer 1

您的 on_pause功能集pause_为真，而 check_pause 中的谓词验证它是否设置为 false。因此调用 notify_all在 on_pause毫无意义，因为 check_pause 中的通知线程将检查谓词并立即返回 sleep 状态。自从pause_是原子的，你不需要调用 notify_all ，你也不需要锁。