Exit infinite while loop (python) when user enters "exit", but if users enters a number compare it to 50

Question

i need help with this problem. ask user to enter a number inside an infinite while loop, compare this number to let say 50 (less, greater or equal). but in order to exit this while loop user must enter "exit". i have the following code that is working according to requirement but i want 'exit' (if) statement to be written at the end. doing so definitely causes an error. please feel free for alternative ways.

while True:
x = input('please enter a number to compare or enter "exit" to exit the loop \n')
if x == "exit":
    exit()
elif int(x) > 50:
    print(x, 'is greater than 50')
elif int(x) < 50:
    print(x, 'is less than 50')
else:
    print('the number you entered is 50')

There are lots of inputs here that will cause errors as well; the only non-integer argument you filter out before calling `int(x)` is the string `exit`. — chepner, Oct 17 '18 at 15:23
You can't put the `exit` case at the end, because you have to ensure that `"exit"` isn't passed to `int`. — chepner, Oct 17 '18 at 15:24
If you use `break` instead of `exit()` this code runs fine, barring other string input than 'exit' — G. Anderson, Oct 17 '18 at 15:28
@G.Anderson yes it works that way but what can you do when your instructor wants you to explicitly ask to enter exit :P, — jahanzaib younis, Oct 17 '18 at 15:43
@chepner yes that's right, i got this question in an assignment and it was explicitly asked to enter 'exit' to leave the loop, other than that it is working fine with try and except — jahanzaib younis, Oct 17 '18 at 15:47
@G.Anderson no problem dude, i just ask for suggestions and all of them are valuable including yours — jahanzaib younis, Oct 17 '18 at 15:48

wizzwizz4 · Accepted Answer · 2018-10-17T15:45:13.980

1

Well, what happens if the user types fkljhae? A ValueError is raised. And... hang on! This is raised for any non-int input - "exit" satisfies this criterion.

from sys import exit

while True:
    x = input('please enter a number to compare or enter "exit" to exit the loop \n')
    try:
        if int(x) > 50:
            print(x, 'is greater than 50')
        elif int(x) < 50:
            print(x, 'is less than 50')
        else:
            print('the number you entered is 50')
    except ValueError:
        if x == "exit":
            exit()

This isn't particularly good, though; what if print raises a ValueError? Let's refactor it so that only int(x) is in the try: except: block:

from sys import exit

while True:
    text = input('please enter a number to compare or enter "exit" to exit the loop \n')
    try:
        x = int(text)
    except ValueError:
        if text == "exit":
            exit()
    else:
        if x > 50:
            print(x, 'is greater than 50')
        elif x < 50:
            print(x, 'is less than 50')
        else:
            print('the number you entered is 50')

This is better, though "exit" isn't at the bottom any more.

edited Oct 17 '18 at 15:45

answered Oct 17 '18 at 15:25

wizzwizz4

6,140
2
26
62

yes it works with try and except but if there is any other string then it shows no error and i doesn't exit, i guess that's out of requirement as i mention explicitly to enter 'exit' to leave the loop – jahanzaib younis Oct 17 '18 at 15:35
I'd refactor a bit. There's a lot of code in the `try` block, and the only thing that can raise the exception you are interested in is the call to `int`. (Granted, there's nothing else that *can* raise a `ValueError` in this case, so you aren't masking any other exceptions, but in principle, you want to keep the `try` block as focused as possible.) – chepner Oct 17 '18 at 15:39
@chepner Correct. However, that doesn't put `"exit"` last... I'll post a better version, too. – wizzwizz4 Oct 17 '18 at 15:40
@jahanzaibyounis You can only accept one answer; the "tick" should go to the answer that answers the question best. You can upvote multiple answers. (Just so you know; I noticed that you clicked both. It's fine to accept another answer, if you feel that it answered your question better.) – wizzwizz4 Oct 17 '18 at 15:41
@wizzwizz4 that was a mistake i clicked both, your answer is much more explained and surely helped me with my problem. – jahanzaib younis Oct 17 '18 at 16:39

score 0 · Answer 2 · answered Oct 17 '18 at 15:23

0

It is because you are trying to parse "exit" which is string to int You can use try and except.

answered Oct 17 '18 at 15:23

kaan bobac

729
4
8

Exit infinite while loop (python) when user enters "exit", but if users enters a number compare it to 50

2 Answers2