Finding the name of a function while defining it

Chris Rebert clp2 at rebertia.com
Thu Dec 27 09:58:27 CET 2012


On Dec 26, 2012 11:55 PM, "Abhas Bhattacharya" <abhasbhattacharya2 at gmail.com>
wrote:
>
> On Thursday, 27 December 2012 10:22:15 UTC+5:30, Tim Roberts  wrote:
> > Abhas Bhattacharya <abhasbhattacharya2 at gmail.com> wrote:
> >
[Oh god please stop/avoid using Google Groups with its godawful
reply-quoting style that adds excessive blank lines]
> > >While I am defining a function, how can I access the name (separately
as
> > >string as well as object) of the function without explicitly naming
> > >it(hard-coding the name)?
> > >For eg. I am writing like:
> >
> > >def abc():
> > >    #how do i access the function abc here without hard-coding the
name?
> >
> > Why?  Of what value would that be?
<snip>
> Because I have this situation:
> I have used a dictionary with "function_name":value pair in the top of
the code. Now when some function is called, I need to print the value
assigned to its name in the dictionary (the functions are defined after the
dictionary). Now there is only one bad way-around for me: I need to
hard-code the name in the function like this:
> def function_name():
>     print(dict_name.get("function_name"))
> but ofcourse it is a bad thing to do because I have a lot of this type of
 functions. It would be better if I can can use the same code for all of
them, because they are all essentially doing the same thing.

I agree with the general outline of Mitya's suggestion, i.e. refactor the
"print the associated value" step into a separate function, thus obviating
the self-reference issue; it'd be bad to repeat that code in each function
anyway.

Anyhow, here's a simple variation that exploits decorators (because they're
generally awesome & one of my favorite features):

def printing_name_beforehand(func):
    def wrapper(*args, **kwargs):
        print(the_dict.get(func.__name__))
        return func(*args, **kwargs)
    return wrapper

Usage:

@printing_name_beforehand
def some_func(...):
    # whatever

(Forgive me if there are typos; composing this reply on a tablet is
cumbersome.)
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