Eryk Sun <[email protected]> added the comment:
> It fails with "The system cannot find the file
> C:\Users\<username>\AppData\Local\Microsoft\WindowsApps\python3.9.exe."
The shell's CreateProcessW() call failed with ERROR_FILE_NOT_FOUND (2). The
file exists, but it's probably a broken appexec link. I'm attaching a script
that displays the contents of an appexec link. For example, here's the working
"python3.9.exe" link:
C:\>read_appexec.py "%LocalAppData%\Microsoft\WindowsApps\python3.9.exe"
Version: 3
Package ID: PythonSoftwareFoundation.Python.3.9_qbz5n2kfra8p0
Entry Point: PythonSoftwareFoundation.Python.3.9_qbz5n2kfra8p0!Python
App Type: 0
Target Path:
C:\Program
Files\WindowsApps\PythonSoftwareFoundation.Python.3.9_3.9.2288.0_x64__qbz5n2kfra8p0\python3.9.exe
---
> The executable in Program Files cannot be launched except in a few
> oddly specific circumstances.
The SYSTEM, LOCAL SERVICE, and NETWORK SERVICE accounts can directly execute
the real executable. For standard users, however, a conditional access control
entry is set in the file's permissions that allows execute access only if the
accessing security context has the app's WIN://SYSAPPID, such as
"PYTHONSOFTWAREFOUNDATION.PYTHON.3.9_QBZ5N2KFRA8P0". When CreateProcessW()
spawns an app from an appexec link, it adds the required WIN://SYSAPPID to a
new token that gets created for the process. The app itself can thus directly
load and execute binary images (EXEs, DLLs) from its installation directory
under "%ProgramFiles%\WindowsApps".
----------
nosy: +eryksun
Added file: https://bugs.python.org/file50434/read_appexec.py
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