How to use sys.exit() in Python

Question

player_input = '' # This has to be initialized for the loop    
while player_input != 0:    
    player_input = str(input('Roll or quit (r or q)'))   
    if player_input == q: # This will break the loop if the player decides to quit            
        print("Now let's see if I can beat your score of", player)
        break
    
    if player_input != r:    
        print('invalid choice, try again')

    if player_input ==r:    
        roll= randint (1,8)    
        player +=roll #(+= sign helps to keep track of score)    
        print('You rolled is ' + str(roll))    
        if roll ==1:    
            print('You Lose :)')    
            sys.exit    
            break

I am trying to tell the program to exit if roll == 1 but nothing is happening and it just gives me an error message when I try to use sys.exit()

This is the message that it shows when I run the program:

Traceback (most recent call last):
 line 33, in <module>
    sys.exit()
SystemExit

Please try to post a complete code snippet - where is player coming from for instance? I've also re-tagged for Python 3.x — Jon Clements
– Jon Clements, Commented Feb 1, 2013 at 3:14
The problem is that you are running your code into IDLE. IDLE catches all exceptions(even SystemExit) and thus you see that traceback. To see how thing would normally go simply run your python program into a python shell (launch python from the terminal/command prompt) — Bakuriu
– Bakuriu, Commented Feb 1, 2013 at 6:49

godidier · Accepted Answer · 2017-03-03 12:36:12Z

43

I think you can use

sys.exit(0)

You may check it here in the python 2.7 doc:

The optional argument arg can be an integer giving the exit status (defaulting to zero), or another type of object. If it is an integer, zero is considered “successful termination” and any nonzero value is considered “abnormal termination” by shells and the like.

answered Mar 3, 2017 at 12:36

godidier

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Comments

Pedro Fontes · Accepted Answer · 2019-09-10 15:47:39Z

11

you didn't import sys in your code, nor did you close the () when calling the function... try:

import sys
sys.exit()

answered Sep 10, 2019 at 15:47

Pedro Fontes

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Comments

Abhijit · Accepted Answer · 2013-02-01 03:09:51Z

8

sys.exit() raises a SystemExit exception which you are probably assuming as some error. If you want your program not to raise SystemExit but return gracefully, you can wrap your functionality in a function and return from places you are planning to use sys.exit

answered Feb 1, 2013 at 3:09

Abhijit

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Comments

mana · Accepted Answer · 2020-08-31 17:07:22Z

4

In tandem with what Pedro Fontez said a few replies up, you seemed to never call the sys module initially, nor did you manage to stick the required () at the end of sys.exit:

so:

import sys

and when finished:

sys.exit()

answered Aug 31, 2020 at 17:07

mana

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Comments

Jon Clements · Accepted Answer · 2013-02-01 03:20:58Z

3

Using 2.7:

from functools import partial
from random import randint

for roll in iter(partial(randint, 1, 8), 1):
    print 'you rolled: {}'.format(roll)
print 'oops you rolled a 1!'

you rolled: 7
you rolled: 7
you rolled: 8
you rolled: 6
you rolled: 8
you rolled: 5
oops you rolled a 1!

Then change the "oops" print to a raise SystemExit

answered Feb 1, 2013 at 3:20

Jon Clements

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Comments

Stas Sorokin · Accepted Answer · 2022-02-15 08:48:38Z

2

Real-World Example

Option 1: `sys.exit()`

Exits Python and raising the SystemExit exception.

import sys
try:
  sys.exit("This is an exit!")
except SystemExit as message:
  print(message)

Output:

This is an exit!

Option 2: `sys.exit()`

Exits Python without a message

import sys
sys.exit()

# runtime: 0.07s

edited Feb 15, 2022 at 8:48

answered Feb 14, 2022 at 16:13

Stas Sorokin

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1 Comment

tdMJN6B2JtUe Over a year ago

I would never write code that simply tries to not run.

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