How can I get the file name of the function that is passed to my decorator in python?

Question

I want to get the original file/scriptname, etc of the function that is being decorated. How can I do that?

 def decorate(fn): def wrapped(): return "scriptname: " + fn.scriptname? return wrapped

I tried using fn.__code__ but that gave me more stuff that I needed. I could parse that string to get the function name, but was wondering if there is a more elegant way to do it

on a side note relying on this kind of stuff in production code is usually a bad idea — Joran Beasley
– Joran Beasley, Commented Aug 28, 2012 at 4:55

John La Rooy · Accepted Answer · 2012-08-28 05:05:32Z

17

import inspect inspect.getfile(fn)

This won't work for builtin functions though, you have to fall back to inspect.getmodule for those.

edited Aug 28, 2012 at 5:05

answered Aug 28, 2012 at 5:00

John La Rooy

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MostafaR · Accepted Answer · 2012-08-28 04:49:22Z

1

Try this:

return "filename: " + fn.func_code.co_filename

answered Aug 28, 2012 at 4:49

MostafaR

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2 Comments

Matthew Trevor Over a year ago

It's important to note that this is relying on implementation details and may not work outside of CPython. Using inspect is the actual correct answer, but why go for correct when the one-liner is easier, right? :|

MostafaR Over a year ago

@Matthew Trevor Correct, The inspect is better, I didn't know about this and learned it from gnibbler's answer.

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