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.

• Psytho: Your if-branch does nothing so you can throw it out. And you can break a loop with 'break'.
• Yotam Salmon: I liked that previous title. Added some 'mystery' to the question ;)

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.

• Ben: What's the point in the looping variable? Why not while True: and break? I wouldn't say your solution is 'typical Python'
• Yotam Salmon: There is a mistake in your answer. A or B will return A not if B is false, but if A is not false. Also, while True: and break are a good solution, mainly in Python, and it's completely acceptable.
• Alexis Clarembeau: I fixed my mistake. Thank you. For the break statement, my teachers always taught me not to use it too often so I never use it. But, I think in that case, it's totally acceptable. It's a very discussed point where to use and not to use break instructions.
• FearlessENT: ok all I had to do was change the if else statement and it worked. ty I spent ages. also your way of looping is very interesting I may use that :D ty once again
• Yotam Salmon: @AlexisClarembeau My teacher tells us to never use 'break', but who really listens? An awesome command that is not used it a shame, isn't it? :)

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.

Answer 3

Well, there is a problem with this line: ans1 == ("Y" or "YES")

It is not similar to this line:

ans1 == "Y" or ans1 == "YES"

The second one is right, the first one is called null coalescing. That's not what you want. Basically, the idiom x or y returns x if x is not null, otherwise it returns y.

So basically you just check if ans1 is "Y" (not "YES")

You can check if a list of idioms contains yours this way:

if ans1 in ["Y", "YES"]:

And you can continue adding values to that list as many as you want.

Answer 4

You can use list and in statement:

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 in ["Y","YES"]:
 x = x
 elif ans1 in ["N","NO"]:
 x = x + 10
print(x)

This ensures that the code works.