Looking up caller's namespace

[andres at corrada.com]

On Thu, May 25, 2000 at 06:42:38PM +0900, Yves-Eric Martin wrote:
>
>     I need your help for a problem my little Python knowledge cannot
> solve. In short, what I want to do is to access, from a function in a
> module, an object of the caller. The catch is that I don't want to pass
> it as an argument to the function, but to look it up in the caller's
> namespace. (See note 2 for why I want to do that).
>. 
>.
>. 
> Note 1: I am currently using the following workaround (which may actually
>     be nicer than what I am trying to do, but it does not answer my
>     question about accessing caller's namespace):
> 
> --- mymodule.py ---
> def myadd(y,z):
>     return y+z
> -------------------
> 
> ----- test.py -----
> import mymodule 
> def myadd(a):
>     return mymodule.myadd(a,b)
> a,b=3,4
> print myadd(a)
> -------------------
> 

I don't know the answer to your question but I came up with another
workaround:

class AllKnowing:

    def __init__( self, callingNamespace ):
        self.__dict__ = callingNamespace
	
    def sum( self, a ):
        z = self.__dict__['b']
	return a + z
	
Firing up the interpreter I get:

>>> knowItAll = AllKnowing( globals() ) 
>>> b = 1 
>>> knowItAll.sum( 2 ) 
3  

------------------------------------------------------
Andres Corrada-Emmanuel   Email: andres at corrada.com
------------------------------------------------------