How to programatically detect if code is running in nuitka compiled or python interpreted mode

Question

Can anyone say how to detect if code is running in an exe created by Nuitka or in a normal python interpreter?

I think I would ideally like an "is_nuitka" flag that would be set to True when compiled and presumably not exist at all when not compiled.

Could then use code like this:

if '__is_nuitka__' in locals() or '__is_nuitka__' in globals():
    print('debug info: running in nuitka mode')

Any suggestions? Is there anything like this available? any alternative approaches?

I am looking for a solution for that question as well. Did you find a solutin? — Stefan, Apr 10 '18 at 20:21

Tiger-222 · Answer 1 · 2019-04-07T09:01:07.387

7

Starting with Nuitka 0.6.2, you can use this code:

is_nuitka = "__compiled__" in globals()

edited Apr 07 '19 at 09:01

answered Apr 07 '19 at 06:50

Tiger-222

I wish there were something like sys.pypy_version_info for nuitka. Checking to see if something is compiled isn't necessarily going to continue working as more Python compilers are created. – dstromberg Jun 19 '21 at 23:28

score -2 · Answer 2 · answered Jan 16 '17 at 15:44

-2

The first idea is to check sys.argv[0] like

if '.exe' in sys.argv[0]:
    # compiled

answered Jan 16 '17 at 15:44

Pavel M.

`python my_script_to_modify_.exe_files.py` would already break your assumptions. – RedX Jan 16 '17 at 15:48
1

if sys.argv[0].endswith('.exe') then =) – Pavel M. Jan 16 '17 at 15:50
WIth Nuitka 0.6.15.3 on a Linux system, the file extension is .bin, not .exe. It would better to get something from the sys module - if nuitka had that. – dstromberg Jun 21 '21 at 21:59

2 Answers2