On Mon, 2008-03-24 at 03:19 -0700, Julien wrote: > Hi all, > > This is more a Python issue than a Django one, but I thought the > Python-aware people that you are might be able to help me :) > > I would like to do something like: > > def called(arg) > if arg==True: > !!magic!!caller.return 1 > > def caller(arg) > called(arg) > return 2
We seem to be turning into comp.lang.python here. :-( Sounds like you want to use an exception and have caller catch an exception of a particular type and then return the value in the exception object. [...] > def called(arg) > if arg==True: > caller_frame = inspect.stack()[1] > ... > > Here 'caller_frame' contains the frame of the caller function. Now, > how can I make that frame return a particular value? That would be incredibly fragile and very poor programming practice, so I'm not going to encourage it by providing an answer. Bytecode level changes are considered hacks for good reason. Malcolm -- How many of you believe in telekinesis? Raise my hand... http://www.pointy-stick.com/blog/ --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
