Python While loop won't exit when asked to

Question

I don't understand why when the user enters "0" the loop won't exit.

def floatInput():
    done = False
    while not done:
        integerIn = input("Please enter an integer < 0 to finish >: ")
        try:
            integerIn = int(integerIn)            
        except:
            print("I was expecting an integer number, please try again...")
            import sys
            sys.exit()
        if integerIn == "0":
           done = True
        else:
            integers.append(integerIn)
    return integers

Pax Vobiscum · Accepted Answer · 2016-04-13 18:42:45Z

2

def floatInput():
    done = False
    while not done:
        integerIn = input("Please enter an integer < 0 to finish >: ")
        try:
            integerIn = int(integerIn)            
        except:
            print("I was expecting an integer number, please try again...")
            import sys
            sys.exit()

Everything above here is fine, but as soon as you got to the comparison, you forgot that you've casted the input to an int.

if integerIn == "0":

Should be

if integerIn == 0:

edited Apr 13, 2016 at 18:42

Pax Vobiscum

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answered Apr 13, 2016 at 17:57

unknown_user_

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Comments

RomCoo · Accepted Answer · 2016-04-13 18:18:02Z

2

The reason is because integerIn is an integer and you are treating it like a string in if integerIn=="0". Replace it with integerIN==0 will do the job.

edited Apr 13, 2016 at 18:18

RomCoo

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answered Apr 13, 2016 at 17:55

Gonguela

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Comments

Pythonista · Accepted Answer · 2016-04-13 17:48:23Z

1

You're converting to an integer and then checking for equality with the string "0".

answered Apr 13, 2016 at 17:48

Pythonista

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Comments

Eekan · Accepted Answer · 2016-04-13 18:27:37Z

EDIT: screw the advice about using input or raw_input. Just saw you python 3.x tag, but decided to leave it for future readers.

You have few problems there...

First, in this line:

integers.append(integerIn)

where is integers to begin with? unless it's a global name you must define it in your function.

Second, in this line:

if integerIn == "0":

you're comparing integer to string here, and here's a thing: in python (using python 2.7 here) a string will be bigger than any number if you're doing a comparison, so integerIn == "0" will evaluate to False, always.

Fix it with this:

if integerIn == 0:

Finally, I should tell you this... your code the way it looks like will throws NameError instead of executing what you've done in your except statement.

Try it with the following test cases and try to explain the behavior yourself :)

Please enter an integer < 0 to finish >: test
Please enter an integer < 0 to finish >: "test"

To avoid such problem next time, use raw_input instead of input. So this line:

integerIn = input("Please enter an integer < 0 to finish >: ")

should be like this:

integerIn = raw_input("Please enter an integer < 0 to finish >: ")

NOTICE: I'm not sure but I think raw_input doesn't exist in python 3.x, instead input there will do exactly the same. please correct if I'm wrong. However, If you're using python 3 then I think you should have no problem.

Here's input vs raw_input() in python 2.x:

input will evaluate the user input then return it.

raw_input will return the user input as string.

so:

# python 2.x
foo = input("input something")     # input 3 + 5
print foo                          # prints 8

bar = raw_input("input something") # input 3 + 5
print bar                          # prints "3 + 5"

Pax Vobiscum · Accepted Answer · 2016-04-13 18:43:18Z

1

Try this

if integerIn == 0:

should work now.

edited Apr 13, 2016 at 18:43

Pax Vobiscum

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answered Apr 13, 2016 at 17:54

Nadeem Ali

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