According to How to exit from Python without traceback?, calling sys.exit()
in a Python script should exit silently without a traceback.
import sys
sys.exit(0)
However, when I launch my script from the command line on Windows 7 with python -i "exit.py"
, (or from Notepad++), a traceback for a SystemExit
exception is displayed.
U:\>python -i "exit.py"
Traceback (most recent call last):
File "exit.py", line 2, in <module>
sys.exit(0)
SystemExit: 0
>>>
Why is sys.exit()
displaying a traceback when run from the Windows command line?
(For reference, I am using Python 3.6.4 on Windows 7)
You're running Python with the
-i
flag.-i
suppresses the usual special handling of theSystemExit
exceptionsys.exit
raises; since the special handling is suppressed, Python performs the normal exception handling, which prints a traceback.Arguably,
-i
should only suppress the "exit" part of the special handling, and not cause a traceback to be printed. You could raise a bug report; I didn't see any existing, related reports.Because you are using the -i option. Try it without that and you won't get a stack trace.
As Matt_G says,
sys.exit(0)
is exactly the same as raisingSystemExit()
, which can be caught in higher level, which in your case is happening since you are using-i
.If you want to exit without traceback, there is
os._exit(0)
which calls a "C function" and exit immediately even in-i
modeas @user2357112 told me,
os._exit(0)
is drastic move that exits without any cleanup. No finally,__exit__
,atexit
,__del__
, etc.Because under the hood,
sys.exit(0)
raises aSystemExit
exception.Further reading here
what you want is:No exception shown:
and your program is terminated.
Run with
-i
option for interactive (inspect interactively after running script) and the exception is shown:because the interpreter keeps running.