Question

I’ve made a Farkle Jabber game bot using Python SleekXMPP library.
In multiplayer mode, a user plays against a user by turns. I’m trying to make a timeout duration so that if your opponent didn’t respond in 1 minute for example, you win.

Here’s what I’ve tried:

import sleekxmpp
...
time_received={}
class FarkleBot(sleekxmpp.ClientXMPP):
    ...
    def timeout(self, msg, opp):                  
        while True:
            if time.time() - time_received[opp] >= 60:
               print "Timeout!"
               #stuff
               break
    def messages(self, msg):
        global time_received
        time_received[user] = time.time()
        ...
        if msg['body'].split()[0] == 'duel':
            opp=msg['body'].split()[1]  #the opponent
            ...   #do stuff and send "Let's duel!" to the opponent.
            checktime=threading.Thread(target=self.timeout(self, msg, opp))
            checktime.start()

The problem with the code above is that it will freeze the whole class until the 1 minute passes. How can I avoid that? I tried putting the timeout funcion outside the class, but nothing’s changed.

Answer 1

If you must wait for something, it is best to use time.sleep() instead of busy waiting. You should rewrite your timeout like this:

def timeout(self, msg, opp, turn):
    time.sleep(60)
    if not turn_is_already_done:
        print "Timeout"

As you see, you must somehow keep track of whether a move has been received on time.

Therefore an easier solution might be to set an alarm using threading.Timer. You must then set a handler to handle the timeout. E.g.

def messages(self, msg):
    timer = threading.Timer(60, self.handle_timeout)
    # do other stuff
    # if a move is received in time you can cancel the alarm using:
    timer.cancel()

def handle_timeout(self):
    print "you lose"