En Sun, 10 Jun 2007 18:16:12 -0300, James T. Dennis <[EMAIL PROTECTED]>
escribió:
> When I try something like this I run into a little problem:
>
> class Foo:
> def foo(self):
> return "foo"
> class Bar:
> def bar(self):
> return "bar"
>
> f = Foo()
> f.__dict__.update(Bar.__dict__)
>
> ... the problem is that while f now has a "bar" method it doesn't
> work quite like a normal instance method:
>
> >>> f.bar()
> Traceback (most recent call last):
> File "<stdin>", line 1, in ?
> TypeError: bar() takes exactly 1 argument (0 given)
> >>>
>
> ... though I can get by with f.bar(f)
Bar.__dict__ contains *unbound* methods - that is, methods not linked to
any particular instance. If you copy them directly into f.__dict__ you
lose the "magic" that binds methods to instances.
You could store a bound method into the instance but it's not a good idea
(there are cyclic references).
I think the easiest way is to define a dynamic class (as the subject on
this old thread suggests):
py> f = Foo()
py> Foo2 = type("Foo2", (Foo,Bar), {})
py> f.__class__ = Foo2
py> f.bar()
'bar'
The code above changes the object class once it was created, but you don't
have to. Also, you don't have to use a different class name (altough it
may be confusing...):
py> f = type("Foo", (Foo,Bar), {})()
py> f.bar()
'bar'
> This "new" module seems to be the key to it all;
It's almost useless now that types are callable.
> but the only docs I have for that say:
>
>>>> help(new)
> Help on module new:
> [...]
> MODULE DOCS
> http://www.python.org/doc/current/lib/module-new.html
Did you follow the above link?
--
Gabriel Genellina
--
http://mail.python.org/mailman/listinfo/python-list
