How to exit from Python without traceback?

You are presumably encountering an exception and the program is exiting because of this (with a traceback). The first thing to do therefore is to catch that exception, before exiting cleanly (maybe with a message, example given).

Try something like this in your main routine:

import sys, tracebackdef main():    try:        do main program stuff here        ....    except KeyboardInterrupt:        print "Shutdown requested...exiting"    except Exception:        traceback.print_exc(file=sys.stdout)    sys.exit(0)if __name__ == "__main__":    main()

Perhaps you're trying to catch all exceptions and this is catching the SystemExit exception raised by sys.exit()?

import systry:    sys.exit(1) # Or something that calls sys.exit()except SystemExit as e:    sys.exit(e)except:    # Cleanup and reraise. This will print a backtrace.    # (Insert your cleanup code here.)    raise

In general, using except: without naming an exception is a bad idea. You'll catch all kinds of stuff you don't want to catch -- like SystemExit -- and it can also mask your own programming errors. My example above is silly, unless you're doing something in terms of cleanup. You could replace it with:

import syssys.exit(1) # Or something that calls sys.exit().

If you need to exit without raising SystemExit:

import osos._exit(1)

I do this, in code that runs under unittest and calls fork(). Unittest gets when the forked process raises SystemExit. This is definitely a corner case!

import syssys.exit(1)