How to get the package, file, and line of a method/function invocant?

Marc 'BlackJack' Rintsch bj_666 at gmx.net
Tue Sep 12 08:33:15 CEST 2006


In <1158038477.221439.15260 at e63g2000cwd.googlegroups.com>, metaperl wrote:

> # Of course I could cheat and pass it, but I don't want to:
> 
> directories = data.storage.logic(__file__)

Why do you consider a plain and simple solution cheating?

Ciao,
	Marc 'BlackJack' Rintsch



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