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.

Your if-branch does nothing so you can throw it out. And you can break a loop with 'break'. — Psytho
– Psytho, Commented Jun 4, 2016 at 20:42
I liked that previous title. Added some 'mystery' to the question ;) — Yotam Salmon
– Yotam Salmon, Commented Jun 4, 2016 at 20:46

Alexis Clarembeau · Accepted Answer · 2016-06-04 20:48:29Z

6

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.

edited Jun 4, 2016 at 20:48

answered Jun 4, 2016 at 20:38

Alexis Clarembeau

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7 Comments

Ben Over a year ago

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 Over a year ago

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 Over a year ago

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 Over a year ago

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 Over a year ago

@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? :)

|

Pierre.Sassoulas · Accepted Answer · 2016-06-04 21:21:47Z

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.

edited Jun 4, 2016 at 21:21

answered Jun 4, 2016 at 21:16

Pierre.Sassoulas

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Comments

Yotam Salmon · Accepted Answer · 2016-06-04 20:40:37Z

1

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.

answered Jun 4, 2016 at 20:40

Yotam Salmon

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Mahmoud Salah · Accepted Answer · 2024-02-19 19:52:40Z

0

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.

answered Feb 19, 2024 at 19:52

Mahmoud Salah

1

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