信号量实现-在通知condition_variable之前，notify（）应该释放互斥量吗？

I've seen a lot of implementations of the semaphore where the notify() function looks something like this: (this particular example is from here)

void notify() {
        std::lock_guard<decltype(mutex_)> lock(mutex_);
        ++count_;
        condition_.notify_one();
    }

I don't understand the reason behind holding the lock while calling notify_one(). Even if a spurious wake-up occurs after releasing the mutex but before notifying the condition variable, everything should work fine (since condition_.wait() should use a predicate to handle spurious wakeups anyway).

是否有以下情况：

void notify() {
        {
        std::scoped_lock lock(mutex_);
        ++count_;
        }
        condition_.notify_one();
    }

会失败吗？