Difference between exit() and sys.exit() in Python


Question

In Python, there are two similarly-named functions, exit() and sys.exit(). What's the difference and when should I use one over the other?

1
370
10/7/2016 12:36:16 AM

Accepted Answer

exit is a helper for the interactive shell - sys.exit is intended for use in programs.

The site module (which is imported automatically during startup, except if the -S command-line option is given) adds several constants to the built-in namespace (e.g. exit). They are useful for the interactive interpreter shell and should not be used in programs.


Technically, they do mostly the same: raising SystemExit. sys.exit does so in sysmodule.c:

static PyObject *
sys_exit(PyObject *self, PyObject *args)
{
    PyObject *exit_code = 0;
    if (!PyArg_UnpackTuple(args, "exit", 0, 1, &exit_code))
        return NULL;
    /* Raise SystemExit so callers may catch it or clean up. */
    PyErr_SetObject(PyExc_SystemExit, exit_code);
   return NULL;
}

While exit is defined in site.py and _sitebuiltins.py, respectively.

class Quitter(object):
    def __init__(self, name):
        self.name = name
    def __repr__(self):
        return 'Use %s() or %s to exit' % (self.name, eof)
    def __call__(self, code=None):
        # Shells like IDLE catch the SystemExit, but listen when their
        # stdin wrapper is closed.
        try:
            sys.stdin.close()
        except:
            pass
        raise SystemExit(code)
__builtin__.quit = Quitter('quit')
__builtin__.exit = Quitter('exit')

Note that there is a third exit option, namely os._exit, which exits without calling cleanup handlers, flushing stdio buffers, etc. (and which should normally only be used in the child process after a fork()).

425
6/19/2018 8:51:36 PM

If I use exit() in a code and run it in the shell, it shows a message asking whether I want to kill the program or not. It's really disturbing. See here

But sys.exit() is better in this case. It closes the program and doesn't create any dialogue box.


Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow
Icon