Question: I am a programmer new to python trying to write a simple test1() function that uses arg (or \*arg) to print out how it was called using the "name" of the argument passed to it, not the content of the list or any other variable that I am passing. 

The following example:

```
def test1(arg):

    print "name of arg is %r" % arg

alphabet = ['a', 'b', 'c']

test1(alphabet) # prints name of arg is ['a', 'b', 'c']

```

I want it to print out: name of arg is alphabet
I researched and tried several things related to using \*argv but did not succeed. Can someone shed some light on this? I feel like I'm missing something obvious.
Thanks!

Answer: 
This is more than a little ugly, but it's one way to achieve what you're after:

```
import traceback
import re

alphabet = ['a', 'b', 'c']

def test(arg):
    stack = traceback.extract_stack()
    arg_name = re.sub('test\((.*)\)', '\g<1>', stack[-2][-1])
    print arg_name

test(alphabet)
test('foo')

```

The result of which is

```
alphabet
'foo'

```
I'm not quite sure, why you want this, anyway ...

If you want the name of argument available in a function, you have to pass the name of the argument to the function. Python is *call by reference* in all cases.

In your function, `arg` will always be named `arg` inside the function. Its outside variable name is not available inside the function.

```
def test1(arg):
    print "name of arg is %r" % arg

```

However python has variable keyword arguments and there lies a possible solution to your problem. Let me redefine `test1()`:

```
def test1(**kwargs):
   param_name = iter(kwargs).next()
   print "name of parameter is %s" % param_name

```

You can call this function with a keyword argument:

```
alphabet = ['a', 'b', 'c']
test1(alphabet=alphabet)

```

Note, however, that what get's printed is not the variable name from *outside* the function, but the name of the keyword.