Python newbie - Understanding class functions

Question

If you take the following simple class:

class AltString:

    def __init__(self, str = "", size = 0):
        self._contents = str
        self._size = size
        self._list = [str]

    def append(self, str):
        self._list.append(str)

    def output(self):
        return "".join(self._list)

And I successfully invoke the class instance using:

as = AltString("String1")

as.append("String2")

as.append("String3")

When I then invoke the output function using as.output instead of a string being returned, I get the following instead:

unbound method AltString.output

if I call it using as.output() I get the following error:

TypeError: unbound method output() must be called with
  AltString instance as first argument (got nothing instead)

What I am not doing right?

Well, first of all you have no class methods in your code. And your code works just fine. So what you are doing wrong it most likely to not give us the actual code. ;) — Lennart Regebro
– Lennart Regebro, Commented Oct 14, 2009 at 13:54
This isn't ruby: You can't call a method by using "as.output". You must put the parenthesis. — Pod
– Pod, Commented Oct 14, 2009 at 14:28
I also doubt that the code posted is the actual code you actually ran. I doubt you did as.output(). The error message is probably from doing AltString.output(). — S.Lott
– S.Lott, Commented Oct 14, 2009 at 14:42

SilentGhost · Accepted Answer · 2009-10-14 14:01:48Z

8

as is a bad variable name, it is reserved keyword in Python. don't name your variables like this. once you fix it, everything else will be alright. of course you should be doing:

alt_str.output()

edit: I was able to replicate your error messages when trying to apply output to the class: AltString.output, then: AltString.output(). You should be applying the method to the instance of the class instead.

alt_str = AltString('spam')
alt_str.output()

edited Oct 14, 2009 at 14:01

answered Oct 14, 2009 at 13:46

SilentGhost

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Comments

Corey Goldberg · Accepted Answer · 2009-10-14 13:47:29Z

1

'as' and 'str' are keywords, don't shadow them by defining variables with the same name.

answered Oct 14, 2009 at 13:47

Corey Goldberg

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1 Comment

SilentGhost Over a year ago

str is not a keyword, it's just a built-in type. and while it's a bad practice to shadow the built-in, it wouldn't cause such problems.

Stefano Borini · Accepted Answer · 2009-10-14 13:50:27Z

1

Your example is confirmed to work as you expect in python 2.4

>>> from x import *
>>> as = AltString("String1")
>>> as.append("bubu")
>>> 
>>> as.output()
'String1bubu'

In python 2.5 it should also work, but will raise a warning about the use of as, which will become a reserved keyword in python 2.6.

I don't really understand why you obtain such error messages. If you are using python 2.6 it should probably produce a syntax error.

answered Oct 14, 2009 at 13:50

Stefano Borini

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1 Comment

S.Lott Over a year ago

I think that the code shown in the question is not the code actually run.

Bite code · Accepted Answer · 2009-10-14 13:52:12Z

0

I ran the following code :

class AltString:

    def __init__(self, str = "", size = 0):
        self._contents = str
        self._size = size
        self._list = [str]

    def append(self, str):
        self._list.append(str)

    def output(self):
        return "".join(self._list)


a = AltString("String1")

a.append("String2")

a.append("String3")


print a.output()

And it worked perfectly. The only flow I can see is that you use "as", which is a reserved keyword.

answered Oct 14, 2009 at 13:52

Bite code

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2 Comments

Trevor Over a year ago

The use of 'as' was a typo, as when I did try to use it, I too got a syntax error (I've been using Python 2.5.4). When I re-tried again using your example, everything seemed perfectly fine. Not sure what I was doing wrong before, but it seems to be fine now. :)

S.Lott Over a year ago

@Trevor: Please do not provide additional information as a comment on an answer. Please update the question with additional information.

Keeper · Accepted Answer · 2009-10-14 13:52:20Z

0

Just tried your code in Python 2.6.2 and the line

as = AltString("String1")

doesn't work because "as" is a reserved keyword (see here) but if I use another name it works perfectly.

answered Oct 14, 2009 at 13:52

Keeper

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