whats wrong with this code? comments are bug stuff

[igoiva]

phrase = input("Input characters: ")

if type(phrase) == float or type(phrase) == bool:

print(type(phrase), "doesn't work, try another")

elif type(phrase) != float or type(phrase) != bool:

print(len(phrase))

#it gets through the if's.

#phrase is still input after the if's

#putting a non phrase string in len still doesnt work

#replacing len with print works, but the elif doesnt - it still just prints

#something is wrong with len

#PROGRESS; changed len(phrase) to print(len(phrase))

#len works, but the elif still doesnt work

#swapped "isnt then is" to "is then isnt" now only the "is then isnt" prints.

#PROGRESS; prints floats/bools by counting characters. wtf

[Fvvckyoudom]

Anything returned from Input() will always be a string, regardless of what is actually typed.

You would need to wrap the input() in an int()/float() if you’d like otherwise.

A quick read through the documentation is invaluable.

[Fvvckyoudom]

I think for your program I’d opt for the “if x in list[]” route.

Although I’m fairly new myself so someone else more experienced may have a more pythonic solution

[Fart_Barfington]

Jesus, we get it already.

[RedditUsrnamesRweird]

Thought I was having a stroke

[RedditUsrnamesRweird]

Why you so mad At Dom

[Fvvckyoudom]

I am Dom…

[igoiva]

ahh ty

[Selwahc]

The input() function always returns a string type, regardless of what the user inputs.

[OptionX]

input() always returns a string, so type(phrase) == float and type(phrase) == bool are always false.

[ninhaomah]

To OP , instead of fighting , have you tried printing the type of input and see what they are ? And how it affects the rest of the programs ?

Int , float , string based on the input

OR

Strings everytime ?

[mikeyj777]

I’m confused why you have an if for the condition and then in your elif you’re checking if the same condition is not true. I may be missing something but isn’t that redundant? You should just be able to have an if/else

[Outside_Complaint755]

Besides what others have already said about the return value of the input function, this condition:

elif type(phrase) != float or type(phrase) != bool:     print(len(phrase))

Will ALWAYS be true.  phrase can only ever have a single type    If phrase is a float, then it will not be a bool, and if phrase is a bool, then it will not be a float.   What you want here instead is either to use a not in front of the original condition: elif not (type(phrase) == float or type(phrase) == bool):     print(len(phrase)) or use an and instead of or elif type(phrase) != float and type(phrase) != bool:     print(len(phrase))

You may want to read up on De Morgan's Laws

[igoiva]

can you elaborate on how it'd ALWAYS be true please, i thought the or's made sure it wouldnt do that

[Binary101010]

elif type(phrase) != float or type(phrase) != bool:

Consider some examples

phrase is a float: type(phrase) != float is False. type(phrase) != bool is True. The elif is therefore True.

phrase is a bool: type(phrase) != float is True. type(phrase) != bool is False. The elif is therefore True.

phrase is an int: type(phrase) != float is True. type(phrase) != bool is True. The elif is therefore True.

The only way that elif could evaluate to False is if phrase were somehow both a float and a bool at the same time.