On Thursday, 27 December 2012 10:22:15 UTC+5:30, Tim Roberts wrote:
> Abhas Bhattacharya <[email protected]> wrote:
>
> >
>
> >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?
>
>
>
> Note that I'm not merely being obstructionist here. What you're asking
>
> here is not something that a Python programmer would normally ask. The
>
> compiled code in a function, for example, exists as an object without a
>
> name. That unnamed object can be bound to one or more function names, but
>
> the code doesn't know that. Example:
>
>
>
> def one():
>
> print( "Here's one" )
>
>
>
> two = one
>
>
>
> That creates one function object, bound to two names. What name would you
>
> expect to grab inside the function?
>
>
>
> Even more obscure:
>
>
>
> two = lamba : "one"
>
> one = two
>
>
>
> Which one of these is the "name" of the function?
>
> --
>
> Tim Roberts, [email protected]
>
> Providenza & Boekelheide, Inc.
It is of quite value to me.
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.
Now, for your questions:
If i call one() and two() respectively, i would like to see "one" and "two".
I dont have much knowledge of lambda functions, neither am i going to use them,
so that's something I cant answer.
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