Calling a constructor by reference

Russell Blau russblau at
Thu Oct 7 15:51:31 CEST 2004

"Edwin Young" <edwin at> wrote in message
news:m37jq2fxdp.fsf at localhost.localdomain...
> Hi,
> I know there must be a way to do this but the correct syntax eludes me.
> I want to write a method which I can pass different types to and have
> it construct them.
> I tried this:
> >>>
> class X:
>     def __init__(self):
>         pass
> def make_thing(type,*args):
>     return type(args)
> x = X()
> y = make_thing(X)
> print x, y
> >>>
> But get 'TypeError: __init__() takes exactly 1 argument (2
> given)'. Presumably X is treated as an unbound instance method?
> Also, the types I want to create inherit from an old-style class in a
> package, so I don't think I can make X a new-style class (if that's
> relevant).
> How should I go about this?

Hint:  what do you see if you run the following function?

>>> def show_args(type, *args):
        print args

>>> show_args()

You just need to change your constructor [__init__()] to accept the correct
number of arguments.

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excite if you really want to reach me.

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