Python编写一个非常基本的警报

Question

我刚刚用python编写了一个非常基本的闹钟，用来在下午2:15有些食物煮熟时提醒我，但我在某个地方读到过时间。睡觉处理其他程序时会浪费时间，我应该使用什么替代？ 我现在的代码，非常基本-

import time
import winsound
time.sleep(370800)
winsound.Beep(1000,10000)

Answer 1

定期检查。下面的代码应该没问题。您可以根据需要调整timeout变量：

timeout = 10
target = time.time() + 370800

while True:
    diff = target - time.time()
    if diff < timeout:
        if diff > 0:  # highly unlikely that diff will go below 0 but just in case
            time.sleep(diff)
        break
    else:
        time.sleep(timeout)

winsound.Beep(1000,10000)