getting name of passed reference

[Steven D'Aprano]

On Mon, 28 Dec 2009 17:27:21 -0800, Joel Davis wrote:

> For posterity, I figured out a solution:
> 
>  > #!/usr/bin/python
> 
>  > import sys
>  > from traceback import extract_stack
> 
>  > varPassed="varName get"
> 
>  > def MyFunc(varPassed):
>  >         try:
>  >                 raise None
>  >         except:
>  >                 frame = sys._getframe(1)
>  >                 print extract_stack(frame,2)[0][3]
> 
> 
>  > MyFunc(varPassed)


Incorrect. Here's a copy-and-paste from an interactive session using that 
code:


>>> import sys
>>> from traceback import extract_stack
>>>
>>> varPassed="varName get"
>>>
>>> def MyFunc(varPassed):
...     try:
...         raise None
...     except:
...         frame = sys._getframe(1)
...         print extract_stack(frame,2)[0][3]
...
>>> MyFunc(varPassed)
None
>>>



> the print statement returns the full function call including parameters
> as they were written in the script (variable names and all)

I'm afraid not. I don't know what testing you did, but it doesn't work as 
you think it works.

Also, I notice that you've called the variable local to the function the 
same name as the variable in the outer scope. What happens when you do 
this?

>>> x = "something else"
>>> MyFunc(x)
None


> at first glance the solution i came up with seems to be in general the
> same as the one presented there,  are there any portability issues
> you're aware of? 

Yes.


sys._getframe is a PRIVATE function, which means it is subject to change 
without notice. You're using an internal function clearly marked as 
private. It is unlikely to even exist at all in other implementations of 
Python (e.g. Jython, IronPython, PyPy, CLPython, etc.), and it might not 
even exist in future versions of CPython.



> also, when can one _not_ get the name?

When the object doesn't have a name at all, or when it has multiple 
names. Which is "the" name?


-- 
Steven