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

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

perform the modification while the lock is held

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

如果改变不需要唤醒线程，像这里的on_pause函数，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 函数将 pause_ 设置为真，而 check_pause 中的谓词验证它是否设置为假。因此在 on_pause 中调用 notify_all 是没有意义的，因为 check_pause 中的通知线程将检查谓词并立即返回休眠。因为 pause_ 是原子的，你不需要调用 notify_all，你也不需要锁。

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

Is it necessary to acquire the lock and notify condition_variable if no thread needs to wake up?

c++

multithreading

locking

condition-variable