访问Python中嵌套函数之外的对象

Question

我正在努力制作一个实际上是闹钟的程序。这比看起来更复杂，但这是有原因的，我不会在这里详述。不过，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。。。我在下面附上了代码摘要。setAlarm之外的任何对"check"或"set"等的调用都永远找不到"alarm"变量。请告诉我我做错了什么——正如你所看到的，我试图将变量声明为全局变量，但仍然没有成功。。。

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!aaaaaaaaaa"
                    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

您已经将alarm创建为局部变量。默认情况下，函数中定义的变量是局部的，而局部的意思正是它听起来的样子——它只存在于函数内部。

您可以通过显式地将其设置为全局变量来修复此问题。这不是一个好主意，但这是最小的变化。只需将语句global alarm添加到setAlarm和每个想要访问它的函数的顶部

一个更好的解决方案是从setAlarm到return alarm。然后，在调用它的代码中，只存储返回值。例如：

def setAlarm():
    # your existing code
    return alarm
def runAlarm():
    alarm = setAlarm()
    # your existing code

现在，runAlarm对同一对象（也称为alarm）有自己的本地引用，因此它可以使用它。