getting name of passed reference

[Joel Davis]

On Dec 29, 2:04 am, Steven D'Aprano
<ste... at REMOVE.THIS.cybersource.com.au> wrote:
> On Mon, 28 Dec 2009 19:28:32 -0800, Joel Davis wrote:
> > my thanks go out to Emile and Mr Hanson for their responses, I think
> > I've found the solution, much shorter as well:
>
> >     > #!/usr/bin/python
>
> >     > import traceback
>
> >     > def testing ( varPassed ):
> >     >         print traceback.extract_stack()[0][3]
>
> >     > testing("123")
>
> > and it seems the traceback module in general seems to have a lot of
> > history to it. This project is for CPython so compatibility with Jython,
> > Iron Python, et al isn't really that important right now. So as far as
> > functionality and compatibility I think I'm set as long as
> > traceback.extract_stack is 3.0 safe.
>
> I'm afraid that one fails again. Do you actually test your solutions
> before telling us you've solved the problem?
>
> >>> import traceback
> >>> def testing ( varPassed ):
>
> ...     print traceback.extract_stack()[0][3]
> ...
>
> >>> testing("123")
> None
> >>> x = "123"
> >>> testing(x)
>
> None
>
> When a "solution" doesn't work under some circumstances (in this case,
> when run in the interactive interpreter) that's a warning that you need
> to understand when and where it will work before using it in production.
> Otherwise, how do you know that it will work under other circumstances?
>
> Or, find an alternative. What are you actually trying to do? "Get the
> name of a passed reference" is a means to an end. What are you expecting
> to do with it?
>
> --
> Steven

Steven I don't know what your issue is, but it works for me. If your
having trouble running the code, then that's your issue, and I would
appreciate it if you would just shut up until you know what the hell
you're talking about. I say that because if you paid the slightest
attention my first solution that you found so woefully inept IS
BASICALLY YOUR FIRST SOLUTION RECONFIGURED.

You also apparently can't read.