> while cal_opt != 9: > menu() > cal_opt = cal() > if cal_opt == 1: > elif cal_opt == 2: ... > elif cal_opt == 7: > else: > print "That's not an option. Try again." > menu() > cal()
Because you don't have a specific check for the exit value (9) the else part gets executed which asks the user for another value. If you put in a test for 9 that does nothing (ie use 'pass') then the while loop will fail on its next attempt and the program will stop. Other options inclde using 'break' to break out of the loop when cal_opt == 9 or using 'continue' to force the while loop to start again. HTH, Alan G Author of the Learn to Program web tutor http://www.freenetpages.co.uk/hp/alan.gauld _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
