Metaclass Question

[Jay Parlar]

On Oct 6, 2004, at 7:23 PM, CarThe answer is *simple*. In fact, it's 
embarassingly simple. Do it
> inside a loop. Really.
>
> for i in range(10):
>     def newfunc(...):
>         ...
>
> ... will create as many copies of newfunc as you want; it will bind
> all to the same name in the local namespace, but each and every
> instance will be different. The func_name attribute will be 'newfunc'
> for all functions created, though; but if you're using 2.4, you can
> rename them later, by changing the func_name attribute, that is now
> writable.
>

But in the example I posted, new_func already is in a loop (for 
func_name in funcs)
On the surface, it looks like it works. The keys in cls.nameReg[name] 
show different references for the new_funcs, but when they're actually 
called, they all run with the same values.

Jay P.
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