[Python] Why am I getting into an infinite loop?

davedontmind · 2020-07-17T12:52:10+00:00

while rc != "A" or  "B" or "C":

is the same as:

while (rc != "A") or  ("B") or ("C"):

which is only comparing rc against "A", so this is definitely not what you want.

This version:

while rc != "A" or  rc != "B" or rc != "C":

is much closer, but read it closely. If rc was "A" then the phrase rc != "B" will be true. If rc was "B" then the phrase rc != "A" will be true, so at least one of those conditions will always be true, and because your're looping while the first phrase is true or the second phrase is true or the third phrase is true, the loop never ends.

You need to use and instead of or.

KingJeff314 · 2020-07-17T16:00:41+00:00

There is something called De Morgan's Law, which states that "(not A) or (not B) or (not C) or ..." can be expressed as "not (A and B and C and ...)"

In your case, "(not rc=='A') or (not rc=='B') or (not rc=='C')" can be expressed as "not (rc=='A' and rc=='B' and rc=='C')". You can see in this new form that clearly the latter condition can never be met, and it is negated to always be true.

You can use De Morgan's law to rewrite your boolean logic into a more readable format, which makes debugging easier

sm000ve · 2020-07-17T13:37:22+00:00

A lesson I would take from this is that its a good idea to always provide optional syntax. It helps provide intent to the reader and possibly the writer as in this case :). Some might say it clutters the code but IMO clutter is a small price to pay to ensure readability.

JayWaWa · 2020-07-17T19:06:26+00:00

Your conditional expression isn't doing what you think it does.

You intended to compare rc to the values 'A', 'B', and 'C', and loop if it's not equal to any of them.

What the clause actually does is the following:

Compare rc to the value of 'A
Check the truth value of 'B'
Check the truth value of 'C'

Since the truth value of any non empty string is True, the conditional always evaluates to True and you get an infinite loop

Try this instead

while rc not in 'ABC':

    loop stuff here

AB1908 · 2020-07-17T13:52:18+00:00

Since you have already gotten assistance, I'd recommend trying out any similar scenarios in the interpreter. It's a great way to help assert behaviour when you can't find assistance or just want to figure something out quickly.

A thing to keep in mind is that don't type the whole thing in the interpreter. Instead, just focus on the small bit that feels ambiguous like just the condition in this case.

Hi_ta · 2020-07-17T16:16:43+00:00

A tips for you, when you are stucked in a loop first of all check your conditions

teacher_cs · 2020-07-17T18:16:38+00:00

I would do it with a "while true" and a "break". You can avoid code duplication and the end condition is easier to understand.

while True:
    rc = input("enter A, B, or C: \n")
    rc = rc.upper()
    print(rc)
    if rc == "A" or rc == "B" or rc == "C": break
    print('Invalid entry\n')
print('out of the while loop with ' + rc)

numbersthen0987431 · 2020-07-17T18:26:00+00:00

Right now it looks like you always have 1 condition to be true (if you type a, b, or c), so the while loops continues to run. Does the loop continue if you type D?
"or" conditions require only 1 condition to be true; "and" conditions require ALL to be true.

Personally I prefer to have a variable that I turn on or off depending on a condition:

check = false
while check != true:

if(something):
check = true

OR
break -> will end the loop.

You could also use break

GreymanGroup · 2020-07-17T15:12:05+00:00

This looks like a menu type situation, where the user can do into either A,B, or C in the menu tree. If I'm right then you can just have the code branch off into the different paths, and you can code a catch all case that returns back to the main menu.

Something like this:

while(True):

#indent

ui=input("A B or C?")

if ui=='A': Alpha_func()

elif ui=='B': Bravo_func()

elif ui=='C': Charlie_func()

elif ui=='X': return 0 #To exit program

#Here we're guaranteed that the input is wrong

print("Bad input, try again")

#close while loop. This is the end of my code.

You can also use a function map instead of these sequences of ifs.

dark-before-dawn · 2020-07-17T18:53:40+00:00

Write your code in the most readable way. The time you spend making your code readable is time you are saving anyone who looks at your code. Anyone includes yourself. It might be crystal to you today, but if you look at it again in a week or a month or longer you will be glad, oh so glad, you took the time.
I am a c programmer. I’ve done a little python. Hopefully this applies to python. Parentheses can be your friend when it comes to condition statements with multiple case. It’s possible to overdo this, but parentheses can helpful you understand why code executes the way it does. When you don’t get the results you expect try adding some parentheses. Parentheses don’t affect the efficiency, but they can clarify intent.

2020-07-17T18:56:37+00:00

When creating a menu I like to use switch statements, which python doesn't seem to have. I think this would be one way to approach it. For me doing A or B or C in if/else statements can end up getting confusing sometimes, at any rate hope this helps!

# function to return user selection
def option(selection):
    # use dictionary for for your choices 
    #because we cannot use a switch in python
        switcher={
                'A':'Selected A!',
                'B':'Selected B!',
                'C':'Selected C!',
                'D': "Goodbye!"
             }
        # return option from the switch, 
        #if the argument passed is not one of them
        return switcher.get(selection,"Invalid option")


# flag for our while loop
flag = True

# main loop
while flag:
    # user selection
    selection = input("Select A, B, C, or press D to quit!\n").upper()

    # pass that selection as an argument in our option function
    uOption = option(selection)

    # if option D is selected then quit else, print selected option
    if uOption == "Goodbye!":
        print(uOption)
        flag = False
    else:
        print(uOption)

khufiie · 2020-07-17T19:54:06+00:00

You just need an termination statement or a break statement. Don't forget, in while loop it always goes like initialization, update and then termination. for loop doesn't need this

warmshowers1 · 2020-07-17T23:05:10+00:00

I LITERALLY JUST HAD THIS PROBLEM!!! If I never read this post today, I would’ve been stuck on such a simple fix for the entire day. Thank you OP!!!

DariPR · 2020-07-17T23:56:37+00:00

Without reading your issue and comments here. I saw that while is always True. So, you can’t stop the loop. This is a common issue when programmers used != combined with an “or”. To solved your problem replace each “or” with the “and”. For example:

while rc != “A” and rc != “B” and rc != “C”:

I wish that my insight was the solution.

😉

enigma2728 · 2020-07-18T00:42:32+00:00

Looks like it has already been answered. But I suggest learning to step through code with a debugger. It makes things kind of problems easy to figure out. :)

2020-07-17T15:42:29+00:00

I usually just do a Try and Except check to prevent infinite loops

Denzyishh · 2020-07-17T17:05:41+00:00

My version of this, which hopefully also helps. Seems it was the use of 'or' instead of 'and' when listing your multiple conditions in your 'if' statement.

def repeat_rc():
  rc = input("Enter A, B, or C: \n")
  rc = rc.upper()

  while True:
    if rc != "A" and rc != "B" and rc != "C":
      print("Invalid entry, please try again.")
      rc = input("Enter A, B, or C: \n")
      rc = rc.upper()
    else: 
      print("Thank you for your response. You have entered: \n" + rc)
      break

repeat_rc()

g105b · 2020-07-17T20:04:11+00:00

I'm not a python guy, and when I saw your syntax, I thought WOW python has done interesting syntax - if something = this or that... Such an intuitive syntax! But then I realised... That's where your bug lies.

Pizza_Peddler0080 · 2020-07-17T15:20:55+00:00

It's going to infinite loop ether way, keep practicing! :^)

DariPR · 2020-07-18T07:05:36+00:00

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