Python while loop condition check for string

Question

In Codeacademy, I ran this simple python program:

choice = raw_input('Enjoying the course? (y/n)')

while choice != 'y' or choice != 'Y' or choice != 'N' or choice != 'n':  # Fill in the condition (before the colon)
    choice = raw_input("Sorry, I didn't catch that. Enter again: ")

I entered y at the console but the loop never exited

So I did it in a different way

choice = raw_input('Enjoying the course? (y/n)')

while True:  # Fill in the condition (before the colon)
    if choice == 'y' or choice == 'Y' or choice == 'N' or choice == 'n':
        break
    choice = raw_input("Sorry, I didn't catch that. Enter again: ")

and this seems to work. No clue as to why

Hint: can you name a value of choice where choice != 'y' or choice != 'Y' evaluates to False? Additionally, De Morgan's laws may be useful to you. — Kevin
– Kevin, Commented Jul 22, 2015 at 14:12

Martijn Pieters · Accepted Answer · 2015-07-22 14:12:48Z

10

You have your logic inverted. Use and instead:

while choice != 'y' and choice != 'Y' and choice != 'N' and choice != 'n':

By using or, typing in Y means choice != 'y' is true, so the other or options no longer matter. or means one of the options must be true, and for any given value of choice, there is always at least one of your != tests that is going to be true.

You could save yourself some typing work by using choice.lower() and test only against y and n, and then use membership testing:

while choice.lower() not in {'n', 'y'}:

answered Jul 22, 2015 at 14:12

Martijn Pieters

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