Question

我有一个问题，我似乎无法独自解决。我正在编写一个小的python脚本，我想知道为什么我的signal.alarm在它返回的函数之后仍然有效。这是代码：

class AlarmException(Exception):
    pass

def alarmHandler(signum, frame):
    raise AlarmException

def startGame():
    import signal
    signal.signal(signal.SIGALRM, alarmHandler)
    signal.alarm(5)
    try:
        # some code...
        return 1
    except AlarmException:
        # some code...
        return -1

def main():
    printHeader()
    keepPlaying = True
    while keepPlaying:
        score = 0
        for level in range(1):
            score += startGame()
        answer = raw_input('Would you like to keep playing ? (Y/N)\n')
        keepPlaying = answer in ('Y', 'y')

所以问题是当我的startGame()函数返回时，SIGALRM仍在倒计时并关闭我的程序。这是追溯：

Would you like to keep playing ? (Y/N)
Traceback (most recent call last):
  File "game.py", line 84, in <module>
    main()
  File "game.py", line 80, in main
    answer = raw_input('Would you like to keep playing ? (Y/N)\n')
  File "game.py", line 7, in alarmHandler
    raise AlarmException
__main__.AlarmException

我如何继续说SIGALRM在其所处的功能退出时停止？

谢谢！

Answer 1

如果要禁用闹钟，请尝试拨打signal.alarm(0)。

在所有可能情况下，它只是在libc中调用alarm()，而man alarm表示alarm(0)“...使当前警报无效并且信号SIGALRM将不会被传递。”< / p>

函数返回时停止SIGALRM

1 个答案: