python - while loop with conditionals

Question

user = input('> ')
while user != ('a' or 'b' or 'c'):
 print(user)
 break

(I know I should not use 'while' here, but just an example to demonstrate my confusions)

In this code, what I want is: if the user enters 'a' or 'b' or 'c', this 'while loop' will not print out the alphabets. However, there is an error, for 'a', it works fine. But for 'b' and 'c', they still get printed out. May I ask why is that?

• m0r7y: Do like this > while user not in ['a', 'b', 'c', 'd'] or you can use tuple while user not in ('a', 'b', 'c', 'd')

Answer 1

You have to break your ('a' or 'b' or 'c') expression:

user = input('> ')
while user != 'a' and user != 'b' and user != 'c':
 print(user)
 break

When python parses ('a' or 'b' or 'c') it is gonna conclude that 'a' is not null, so it is not necessary to continue the expression in the parenthesis and everything inside became simply 'a'

Or, if you want to keep using your more compact way, your option would be:

user = input('> ')
while user not in ('a', 'b', 'c'):
 print(user)
 break

In this case, you are looking if you user exists in the tuple.

Answer 2

You are testing against 3 conditions here.

while user != 'a'
while 'b'
while 'c'

So while user is anything else than 'a', it will print. You could do something like this.

while user != 'a' or user != 'b' or user != 'c':

or

while user not in ['a', 'b', 'c']:

Answer 3

You can simply use this:

while user not in ["a", "b", "c"]:
 print(user) 
 break