Python 3 weird behavior of subprocess.call

Question

Environment:

Windows: 10
Python: 2.7.13 and 3.8.1
default Python launcher: py -3
Default python: 2.7.13

> python -V
> Python 2.7.13

> py -3 -V
> Python 3.8.1

launcher.py:

import subprocess
subprocess.call(['python', '-V'])

1. Test 1: py -3 launcher.py

   Output: python 3.8.1 (HOW!)

2. Test 2: py -2 launcher.py

   Output: Python 2.7.13

The output should be Python 2.7.13 only, even the launcher runs under py 3!
note that, adding shell=True will work, but the idea is not to use it, and if I ran

subprocess.call(['python', 'script_under_py_2.py']) # Will run python 3 with script python 2!

Thanks
Adam

Answer 1

This behavior is due to the inconsistent handling of the PATH environment variable when calling Popen on Windows and Unix.

Windows creates its sub-processes using CreateProcess function. The search path for CreateProcess includes parent process directory, which may be the reason why you are having different binaries executed.

On Windows, PATH is only considered when shell=True is also passed.

You can learn more about the issue here:

• subprocess PATH semantics and portability (notice the "needs patch" stage)