alarm

I have this decorator: def timed_out(timeout): def decorate(f): if not hasattr(signal, "SIGALRM"): return f def handler(signum, frame): raise TimedOutExc() @functools.wraps(f) def new_f(*args, **kwargs): old = signal.signal(signal.SIGALRM, handler) signal.alarm(timeout) try: result = f(*args, **kwargs) finally: signal.signal(signal.SIGALRM, old) signal.alarm(0) return result new_f.func_name = f.func_name return new_f return decorate The code only does anything on linux, though, as on windows, there is no SIGALRM . What would be the simplest way to have this code work in Windows as well? It's

I was wondering if it's possible (and if it is how) to start up my app at a specific time, something like an alarmclock which goes off at a specific time. Let's say I want my app to start up at 8 in the morning, is that feasable ? ninjasense You can do it with AlarmManager, heres a short example. First you need to set the alarm: AlarmManager am = (AlarmManager) con.getSystemService(Context.ALARM_SERVICE); Date futureDate = new Date(new Date().getTime() + 86400000); futureDate.setHours(8); futureDate.setMinutes(0); futureDate.setSeconds(0); Intent intent = new Intent(con, MyAppReciever.class);

I want a service to run all the time in my application. So I want to restart it even if it is force closed by user. There is definitely a way to do it as apps like facebook are doing it.(Its not done using push notification, facebook restarts its service even if internet is off). Any help would be appreciated. Thanks! Mehul Joisar First of all, it is really very bad pattern to run service forcefully against the user's willingness . Anyways, you can restart it by using a BroadcastReceiver which handles the broadcast sent from onDestroy() of your service. StickyService.java public class