create a function that validates if a number is binary and 8 bits in length

Question

def get8bit():
    while True:
        try:
            bNum = int(input('Please enter your 8 bit number'))
        except:
            print('Please only enter numbers 0 and 1')
        return bNum

def valid8bit(x):
    validChoices = [00000000,11111111]
    while x not in validChoices:
        x = int(input('Please only enter an 8 bit number:'))
        if x == validChoices:
         break
    return x

def convertodeci(y):
    y = y ** 2
    return y
    
user = get8bit()
thisValid = valid8bit(user)
userTodeci = convertodeci(thisValid)
print(userTodeci)

I am trying to write a function that both validates that the number the user entered is 0 or 1 AND is 8 bits in length. I know I need to keep looping until it is correct but I cannot get it to validate the input correctly. Please show me what I'm doing wrong! I want to learn :)

Answer 1

Your code contains a lot of complications and mistakes. For one, you want to accept a string like '00000001', but that's not a valid integer representation, so your first test would make it fail, if it didn't contain errors itself.

You then have a user input command in the middle of a function that supposedly checks if a given number is a valid 8-bit integer, which seems like bad design and not part of the problem you're solving.

And finally, you only try to check if what was entered is 8 1's or 8 0's, but that's not what you want to do either. And the check won't work since 00000000 is not a valid string literal (nor a valid integer).

You were probably after something like:

def valid8bit(s):
    return len(s) == 8 and all(ch in '01' for ch in s)

x = input('Enter an 8 digit binary number:')
print(valid8bit(x))

There's many ways to skip a cat, as @Enzo suggested, this is another approach:

def valid8bit(s):
    try:
        return int(s, 2) in range(256)
    except ValueError:
        return False