Redefining __call__ in an instance

[Robert Ferrell]

Jason Mobarak <jmob at nospam__unm.edu> wrote in message news:<x7KdnVMXjpAPq5rdRVn-sw at comcast.com>...
> def firstFunc (s, word='up'):
>    print "foo"
> 
> class callNoWork(object):
>    def __new__ (cls):
>      cls.__call__ = firstFunc
>      return object.__new__(cls)
> 
> callNoWork()()
> 
> # Dunno if you've read this, but it explains this:
> # http://python.org/2.2.1/descrintro.html

Thanks for the pointer.  I had read this a while ago, it didn't all
sink in, and I'd forgotten about it.  That was exactly the info that I
needed.  In particular, it says that __call__ is a static method, so
it is not possible to define that differently for each instance.

The code above was a good reminder for me, but because __call__ is a
static method, any new instantiation redefines __call__ for all
existing instances.

Thanks for the pointer,
-robert