conditional evaluation of source file in python

Question

Say I have a file that's only used in pre-production code

I want to ensure it gets not run in production code- any calls out to it have to fail.

This snippet at the top of the file doesn't work - it breaks the Python grammar, which specifies that return must take place in a function.

if not __debug__:
   return None

What is the best solution here - the one that doesn't involve making a gigantic else, that is. :-)

score 7 · Answer 1 · answered Mar 08 '11 at 01:32

7

if not __debug__:
    raise RuntimeError('This module must not be run in production code.')

answered Mar 08 '11 at 01:32

dappawit

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The kicker is, I want the module to be "empty" when in production mode. Raising an error is not the right answer here. – Paul Nathan Mar 08 '11 at 16:54
Ahh, apparently I misunderstood your question. Then I think the best choice is the gigantic `if`. I would probably use `if __debug__:`, put everything you want in there, and then you don't need an `else`. I realize you want to avoid this sort of thing, however :) – dappawit Mar 08 '11 at 20:29

samplebias · Answer 2 · 2011-03-08T17:16:13.637

4

Maybe split the non-production code out into a module which is conditionally imported from the main code?

if __debug__:
    import non_production
    non_production.main()

Updated: Based on your comment, you might want to look a 3rd-party library pypreprocessor which lets you do C-style preprocessor directives in Python. They provide a debugging example which seems very close to what you're looking for (ignoring inline debug code without requiring indentation).

Copy/pasted from that url:

from pypreprocessor import pypreprocessor
pypreprocessor.parse()
#define debug

#ifdef debug
print('The source is in debug mode')
#else
print('The source is not in debug mode')
#endif

edited Mar 08 '11 at 17:16

answered Mar 08 '11 at 01:30

samplebias

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That's likely to be the solution, but I am hoping to avoid creating two different files for what amounts to an `#ifdef`. – Paul Nathan Mar 08 '11 at 16:55
@Paul Also check out http://code.google.com/p/pypreprocessor/source/browse/examples/debug2production.py. It demonstrates how to use command line arguments to manipulate the #defines when you execute the script (Ie, the defines are stored in a python list and can be changed in code). That should eliminate the need to create multiple source files. Also, if you get a feel for it, there's a way to output the post-processed code do a file if, for some reason, you need a production-only version. Note: I'm the author of pypreprocessor so if you need any more info feel free to chime in. – Evan Plaice Mar 14 '11 at 05:20

score 1 · Answer 3 · answered Mar 08 '11 at 01:34

1

import sys

if not __debug__:
    sys.exit()

Documentation for sys.exit.

answered Mar 08 '11 at 01:34

Gary Kerr

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score 1 · Answer 4 · edited May 23 '17 at 11:55

One way you can do this is to hide all of the stuff in that module in another module that is imported conditionally.

.
├── main.py
├── _test.py
├── test.py

main.py:

import test
print dir(test)

test.py:

if __debug__:
    from _test import *

_test.py:

a = 1
b = 2

Edit:

Just realized your comment in another answer where you said "I am hoping to avoid creating two different files for what amounts to an #ifdef". As shown in another answer, there really isn't any way to do what you want without an if statement.

I've upvoted the answer by samplebias, as I think that answer (plus edit) describes the closest you're going to get without using an if statement.

conditional evaluation of source file in python

4 Answers4