Segfault launching ipykernel on Python 3.12.3 in VS Code Jupyter (works on Python 3.11)
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Assessment
- Difficulty
- 4/5
- Estimated time
- 3-5 days
- Newbie friendliness
- 35/100
Research direction
Reproduce the failure with python3 -X faulthandler -m ipykernel_launcher -f /tmp/test_kernel.json, then compare it with the working Python 3.11 command. Start from ipykernel_launcher.py, ipykernel/kernelapp.py, and the listed native extensions; done means Python 3.12 starts the kernel and creates its connection file without a segmentation fault.
Written by the indexing model from the issue text.
Description
Summary
Summary
A Jupyter kernel startup failure occurs when launching ipykernel under Python 3.12.3 inside the project virtual environment. The process crashes with Segmentation fault during IPKernelApp initialization. The same kernel startup works correctly on Python 3.11.15.
This appears to be a bug in the ipykernel / IPython startup path for Python 3.12 rather than a notebook-specific issue.
Reproduction Steps
-
Open the project directory:
cd /home/alvaritogsg/personalProjects/nuclioShool/nuclio_charlas -
Activate the virtual environment:
. .venv/bin/activate -
Run the kernel launcher directly with faulthandler:
python3 -X faulthandler -m ipykernel_launcher -f /tmp/test_kernel.json -
Observe the crash:
Segmentation fault (core dumped)- No
/tmp/test_kernel.jsonfile is created.
-
Compare with Python 3.11:
python3.11 -X faulthandler -m ipykernel_launcher -f /tmp/test_kernel_py311.json -
The Python 3.11 kernel starts successfully and remains alive.
Actual Behavior
The kernel process crashes immediately with a segmentation fault while initializing the IPython kernel application.
The stack trace includes:
logging/__init__.pyIPython/core/application.pyipykernel/kernelapp.pytraitlets/config/application.pyipykernel_launcher.py
Loaded native extension modules at the time of the crash:
zmq.backend.cython._zmqtornado.speedupspsutil._psutil_linux
VS Code reports:
The error displaed by VS Code was:
Error: El kernel ha muerto. Error: ... Vea Jupyter [log](command:jupyter.viewOutput) para obtener más detalles.
- The execution that is falling is:
~python -m ipykernel_launcher - The check command
python -c "import ipykernel; print(...)"works correctly.
Expected Behavior
The kernel should start normally and create a connection file such as /tmp/test_kernel.json. Jupyter frontends should be able to connect to the kernel without the process dying.
Environment
- OS: Linux (WSL/Remote VS Code)
- VS Code Jupyter extension:
2025.9.1 - Python 3.12.3 virtualenv:
/home/alvaritogsg/personalProjects/nuclioShool/nuclio_charlas/.venv
Package Versions in .venv
ipython 9.12.0ipykernel 7.2.0jupyter-client 8.8.0jupyter-core 5.9.1traitlets 5.14.3tornado 6.5.5pyzmq 27.1.0psutil 7.2.2
Python Version Comparison
- Python 3.12.3: fails
- Python 3.11.15: works
The working Python 3.11 environment was tested using a temporary venv and a similar ipykernel_launcher command.
Additional Notes
python3 -m IPython --helpworks and evenpython3 -m ipykernel_launcher --helpsucceeds, but actual kernel startup still crashes.- The crash appears to happen before any frontend-level Jupyter communication begins, so this is likely not a VS Code issue.
- The failing path is likely in the
ipykernel/IPythoninitialization logic, possibly specific to Python 3.12 in this environment.
- Dominant language
- Python
- Stars
- 734
- Forks
- 411
- Avg merge
- 1d 5h
- Merged PRs (30d)
- 8
Contributor guide
First steps
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