CODEWITHSUNDEEP
pythonpython-3.x
S. Thomas
S. Thomas

Reputation: 33

Python While loop won't exit when asked to

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

Upvotes: 2

Views: 114

Answers (5)

Nadeem Ali
Nadeem Ali

Reputation: 21

Try this

if integerIn == 0:

should work now.

Upvotes: 1

unknown_user_
unknown_user_

Reputation: 69

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:

Upvotes: 2

Eekan
Eekan

Reputation: 71

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"

Upvotes: 1

Gonguela
Gonguela

Reputation: 25

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.

Upvotes: 2

Pythonista
Pythonista

Reputation: 11615

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

Upvotes: 1

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