[Tutor] Calling super in __init__

[Kent Johnson]

On Mon, May 19, 2008 at 3:27 AM, wcyee <wcyeee at gmail.com> wrote:
> Hi - I wrote a custom exception class as follows:
>
> class CustomError(Exception):
>     def __init__(self, msg):
>         super(CustomError, self).__init__(self, msg)

Should be
  super(CustomError, self).__init__(msg)
i.e. don't repeat 'self'.

> According to the documentation, "super() only works with new-style classes".
> Is the problem that Exception is an "old style" class (I'm still learning
> what this means)?

No

> If so, how can I tell when I'm up against an "old style"
> class when using python?

The type of a new-style class is 'type' (unless it has a custom
metaclass). The type of an old-style class is 'classobj':
In [20]: type(Exception)
Out[20]: <type 'type'>

In [21]: class Foo: pass
   ....:

In [22]: type(Foo)
Out[22]: <type 'classobj'>

New-style classes have a __getattribute__() special method which is
missing in old-style classes, so another way to tell is to look for
__getattribute__.

Kent