Python while loop not working as expected

Question

When I type "no" into the input I expected it to add 1 to "x", therefore ending the loop, but what happens is that it ignores it and does not add 1 x. Here is the code.

x = 1
password = ""

while x == 1:        
    # imagine there is some code here which works

    ans1 = input("\n\nTest a new password? ")
    ans1 = ans1.upper()

    print(ans1)

    if ans1 == ("Y" or "YES"):
        x = x
    elif ans1 == ("N" or "NO"):
        x = x + 10

    print(x)

It's the bottom if/elif statement that is not working. It should continue to ask for input again until the user says NO but this isn't working.

Answer 1

You should use or that way.

if ans1 == ("Y" or "YES"):

Can be replaced with:

if ans1 == "Y" or ans1 == "YES": 

Or:

if ans1 in ("Y", "YES"): 

The bug comes from the definition of the or operator. When you do "Y" or "YES", it will return "Y" as A or B is defined to return A if A is not false. Here, A is "Y" which is not a False value. So, it will return A="Y". If you do if a == ("Y" or "YES"):, il will be equivalent to if a == "Y":. Ok it's a bit tricky but it's how python works.

Moreover, your code is very strange. It's a very bad habit to exit a loop like that. Generally, we put a boolean value "looping" that is set to false when we want to leave the loop.

Here's how I would do your loop:

looping = True 
password = "" 

while looping: 

    ans1 = input("\n\nTest a new password? ")

    if ans1.upper() in ("NO", "N"): 
        looping = False

You can also use a construction with an infinite loop (while True:). Then, you call the instruction break to quit the loop.

Answer 2

You could also use "break" or "exit" to go out of the loop or the program. It's also generally better to use a larger condition that goes well in unexpected case (x<=0 or ans1 isn't YES rather than x==0 or ans1 is YES or ans1 is NO).

while True:
  # Code
  if ans1 not in ["Y", "YES"]:
    break # or exit

Then you would have no undefined behavior, and also fewer condition to take care of : if it isn't "YES" or "Y", the program exit.