My program does not stop when i tell it to exit

Question

When i press 5 for quit it comes up as an error but does not work

For some reason

import sys
sys.exit()

does not work

def ask (user) :

    if user == 1:
        user = input('select a file with an ASCII art image ')
        f = open(user, 'r')
        if f.mode == 'r':
            showart = f.read()
            print(showart)
            print("You will be returned to the menu")

    elif user == 2:
        print('hi')

    elif user == 3:
              print ('hi')

    elif user == 4:
            print("hi")

    elif user == 5:
            print('goodbye')
            import sys
            sys.exit()


while (True) :
    print("1 - Display ASCII art ")
    print("4 - convert RLE option ")
    print("5 - Quit ")
    print("   ")
    print("   ")
    print("   ")

    try:
        user=int(input("Select an integer between 1 and 5 : "))

        if (user<5 and user > 1) :
            ask(user)
        else:
                user=int(input("Please enter a number between 1 and 5 : "))
                while (user > 5 or user < 1) :
                    user=int(input("Please enter a number between 1 and 5 : "))

                ask (user)
    except:
        print('***SORRY, YOUR OPTION DID NOT WORK***\n ***PLEASE SELECT ANOTHER NUMBER***')

it prints goodbye but it does not quit the program

`sys.exit` raises `SystemExit` exception. you are catching it inside `try...except` :https://docs.python.org/3/library/sys.html#sys.exit — ymonad, Jul 16 '19 at 09:19
You might find [Asking the user for input until they give a valid response](https://stackoverflow.com/questions/23294658/asking-the-user-for-input-until-they-give-a-valid-response) useful. — martineau, Jul 16 '19 at 09:30
@BenNoimark in your code, which is above, there are definitely indentation problems. (Not relevant to your error, but saying there arent indentation problems is totally wrong) — NemoMeMeliorEst, Jul 16 '19 at 11:30
@NemoMeMeliorEst And I can ensure you there is no indentation problems. This might not be easy to read due to the inconsistent of the indentation, but this runs fine. — BlueSheepToken, Jul 16 '19 at 14:20

BlueSheepToken · Accepted Answer · 2019-07-16T14:17:00.643

Because sys.exit() throws an error, which is caught by the except and runs again due to the while True

As mentionner in the comments, you can simply add the Exception type, as SystemExit is not a subtype of Exception it won't be catch, and then you can exit the program properly

def ask(user):
    if user == 1:
        user = input('select a file with an ASCII art image ')
        f = open(user, 'r')
        if f.mode == 'r':
            showart = f.read()
            print(showart)
            print("You will be returned to the menu")

    elif user == 2:
        print('hi')
        return 0

    elif user == 3:
        print('hi')

    elif user == 4:
        print("hi")


while True:
    print("1 - Display ASCII art ")
    print("4 - convert RLE option ")
    print("5 - Quit ")
    print("   ")
    print("   ")
    print("   ")

    try:
        user = int(input("Select an integer between 1 and 5 : "))

        """
        If user enter value which is 5 then print 'goodbye'
        as you expect and break the while loop. So it's no
        longer running.
        """
        if user == 5:
            print('goodbye')
            break

        """
        5 > user > 0 mean match only 4, 3, 2, 1

        this loop will continue when the user entered an
        integer value which is not belongs to 4, 3, 2, 1.
        Otherwise it goes to else statement and execute
        ask(user) function.
        """
        if not 5 > user > 0:  # [..., -1, 0, 5, 6, ...]
            continue
        else:
            ask(user)
    except Exception:
        print('***SORRY, YOUR OPTION DID NOT WORK***\n ***PLEASE SELECT ANOTHER NUMBER***')

Or use `except Exception:` as a catch-all, SystemExit is not a subtype of Exception, it’s a subtype of the BaseException — Andrew Morozko, Jul 16 '19 at 10:14

score 0 · Answer 2 · answered Jul 16 '19 at 09:25

You are catching sys.exit

Exit from Python. This is implemented by raising the SystemExit exception, so cleanup actions specified by finally clauses of try statements are honored, and it is possible to intercept the exit attempt at an outer level.

https://docs.python.org/3/library/sys.html#sys.exit

score 0 · Answer 3 · answered Jul 16 '19 at 10:41

I just change your code, but I don't know it match with your requirements and explain what I did in the comment of the code.

def ask(user):
    if user == 1:
        user = input('select a file with an ASCII art image ')
        f = open(user, 'r')
        if f.mode == 'r':
            showart = f.read()
            print(showart)
            print("You will be returned to the menu")

    elif user == 2:
        print('hi')
        return 0

    elif user == 3:
        print('hi')

    elif user == 4:
        print("hi")


while True:
    print("1 - Display ASCII art ")
    print("4 - convert RLE option ")
    print("5 - Quit ")
    print("   ")
    print("   ")
    print("   ")

    try:
        user = int(input("Select an integer between 1 and 5 : "))

        """
        If user enter value which is 5 then print 'goodbye'
        as you expect and break the while loop. So it's no
        longer running.
        """
        if user == 5:
            print('goodbye')
            break

        """
        5 > user > 0 mean match only 4, 3, 2, 1

        this loop will continue when the user entered an
        integer value which is not belongs to 4, 3, 2, 1.
        Otherwise it goes to else statement and execute
        ask(user) function.
        """
        if not 5 > user > 0:  # [..., -1, 0, 5, 6, ...]
            continue
        else:
            ask(user)
    except:
        print('***SORRY, YOUR OPTION DID NOT WORK***\n ***PLEASE SELECT ANOTHER NUMBER***')

If you didn't get anything, please comment to this answer and I'll help you.

My program does not stop when i tell it to exit

3 Answers3