How to get the original variable name of variable passed to a function

Question

Is it possible to get the original variable name of a variable passed to a function? E.g.

foobar = "foo"  def func(var):     print var.origname 

So that:

func(foobar) 

Returns:

>>foobar

EDIT:

All I was trying to do was make a function like:

def log(soup):     f = open(varname+'.html', 'w')     print >>f, soup.prettify()     f.close() 

.. and have the function generate the filename from the name of the variable passed to it.

I suppose if it's not possible I'll just have to pass the variable and the variable's name as a string each time.

Answer 1

EDIT: To make it clear, I don't recommend using this AT ALL, it will break, it's a mess, it won't help you in any way, but it's doable for entertainment/education purposes.

You can hack around with the inspect module, I don't recommend that, but you can do it...

import inspect  def foo(a, f, b):     frame = inspect.currentframe()     frame = inspect.getouterframes(frame)[1]     string = inspect.getframeinfo(frame[0]).code_context[0].strip()     args = string[string.find('(') + 1:-1].split(',')          names = []     for i in args:         if i.find('=') != -1:             names.append(i.split('=')[1].strip())                  else:             names.append(i)          print names  def main():     e = 1     c = 2     foo(e, 1000, b = c)  main() 

Output:

['e', '1000', 'c'] 

Answer 2

To add to Michael Mrozek's answer, you can extract the exact parameters versus the full code by:

import re import traceback  def func(var):     stack = traceback.extract_stack()     filename, lineno, function_name, code = stack[-2]     vars_name = re.compile(r'\((.*?)\).*$').search(code).groups()[0]     print vars_name     return  foobar = "foo"  func(foobar)  # PRINTS: foobar