Question

我试图制作一个有效的闹钟程序。它看起来比它看起来要复杂得多，但有理由说，我不会进入这里。但是，我遇到的问题是我在嵌套函数中创建了一个Alarm对象，但无论我做什么，我都无法在该函数之外访问该对象...我＆＃39 ;附上下面的删节代码。任何打电话给＆＃34;检查＆＃34;或＆＃34;设置＆＃34;在setAlarm之外永远找不到＆＃34;警报＆＃34;变量。请让我知道我做错了什么 - 我尝试将变量声明为全局变量，正如您所见，但它仍然无法解决...

谢谢！

class Alarm(threading.Thread):
    def __init__(self, datetime, grace, close):
        super(Alarm, self).__init__()
        self.datetime = datetime
        self.grace = grace
        self.close = close
        self.keep_running = True

    def run(self):
        try:
            while self.keep_running:
                now = datetime.datetime.now()
                if now > self.datetime + datetime.timedelta(minutes=self.grace):
                    print "Oh no! It's %02d:%02d, which is %d minutes past your alarm time of %02d:%02d!" % (now.hour,now.minute,self.grace,self.datetime.hour,self.datetime.minute)
                    print "ALARM NOW!\a\a\a\a\a\a\a\a\a\a"
                    break
            time.sleep(10)
        except:
            return
    def just_die(self):
        self.keep_running = False

def setAlarm():
    print "What time would you like to set the alarm for? (in 00:00 military time please)"
    wakeup = raw_input()
    wakeuphour = int(wakeup[:2])
    wakeupmin = int(wakeup[3:])
    now = datetime.datetime.now()
    if now.hour > wakeuphour or (now.hour == wakeuphour and now.minute > wakeupmin):
        alarmday = now + datetime.timedelta(days=1)
        alarmtime = datetime.datetime(alarmday.year,alarmday.month,alarmday.day,wakeuphour,wakeupmin)
    else:
        alarmtime = datetime.datetime(now.year,now.month,now.day,wakeuphour,wakeupmin)
    close = 15
    grace = 5
    alarm = Alarm(alarmtime, grace, close)
    if alarmtime.day != now.day:
        print "Your alarm is set for %02d:%02d tomorrow." % (alarmtime.hour,alarmtime.minute)
    else:
        print "Your alarm is set for %02d:%02d today." % (alarmtime.hour, alarmtime.minute)

def runAlarm():
    setAlarm()
    alarm.start()
    while True:
        print "You can say 'stop', 'check', 'change', 'set', or 'quit'"
        text = str(raw_input())
        if text == "stop":
            if alarm != 0:
                alarm.just_die()
                alarm = 0
                print "Okay, I've cancelled the alarm."
            else:
                print "There was no alarm to stop..."
        elif text == "check":
            if alarm == 0:
                print "Sorry, you don't have any alarm set. To create an alarm, type 'set'"
            else:
                pass
        elif text == "change":
            pass
        elif text == "set":
            alarm = 0
            setAlarm()
            alarm.start()
        elif text == "quit":
            print "Sure thing. Bye bye!"
            break
        else:
            print "Sorry, I didn't understand that. Please try again."

Answer 1

您已将alarm创建为本地变量。函数中定义的变量默认是局部的，而local本身就是它听起来的样子 - 它只存在于该函数内部。

你可以通过明确地将其作为全局变量来解决这个问题。这不是一个好主意，但它是最小的变化。只需将语句global alarm添加到setAlarm的顶部以及每个想要访问它的函数。

更好的解决方案是来自return alarm的{{1}}。然后，在调用它的代码中，只存储返回值。例如：

setAlarm

现在，def setAlarm(): # your existing code return alarm def runAlarm(): alarm = setAlarm() # your existing code对同一个对象（也称为runAlarm）有自己的本地引用，因此它可以使用它。

在Python中访问嵌套函数之外的对象

1 个答案: