hi, any help in understanding the program will be highly appreciated. as per the stackless documentation stackless.run() is must to start a scheduler. then how come the below program runs the tasklet "fun" without stackless.run() call?
thanks in advance. On Fri, Jun 19, 2015 at 2:36 AM, temp sha <[email protected]> wrote: > Hi, > > I could not understand how the below program executes function "fun" > without calling stackless.run() in the program? Here "fun" runs as a > tasklet and as per my knowledge for that stackless.run() is must. > > > > ----------------------------------------------------------------- > import stackless > > class A: > def __init__(self,name): > self.name = name > self.ch = stackless.channel() > stackless.tasklet(self.fun)() > > def __call__(self,val): > self.ch.send(val) > > def fun(self): > while 1: > v = self.ch.receive() > print "hi" , v > > > if __name__ == "__main__": > obj = A("sh") > obj(6) > ----------------------------------------------------------------- > > output: > ---------- > hi 6 > > > > > > thanks, > ravi _______________________________________________ Stackless mailing list [email protected] http://www.stackless.com/mailman/listinfo/stackless
