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Mike Lezhnin
Mike Lezhnin

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Python segfault one-liner

Today I found a way to segfault Python, and after some investigation made a minimal working example:

from threading import Thread
from itertools import count

Thread(target = lambda: [print('', end = '') for x in count()]).start()

The program runs infinitely, so you have to press ^C to stop it, and it raises:

Exception ignored in: <module 'threading' from '/usr/lib/python3.6/'>
Traceback (most recent call last):
  File "/usr/lib/python3.6/", line 1294, in _shutdown
  File "/usr/lib/python3.6/", line 1056, in join
  File "/usr/lib/python3.6/", line 1072, in _wait_for_tstate_lock
    elif lock.acquire(block, timeout):
Fatal Python error: could not acquire lock for <_io.BufferedWriter name='<stdout>'> at interpreter shutdown, possibly due to daemon threads

Thread 0x00007f07e3866700 (most recent call first):
  File "", line 4 in <listcomp>
  File "", line 4 in <lambda>
  File "/usr/lib/python3.6/", line 864 in run
  File "/usr/lib/python3.6/", line 916 in _bootstrap_inner
  File "/usr/lib/python3.6/", line 884 in _bootstrap

Current thread 0x00007f07e53a0740 (most recent call first):
Aborted (core dumped)

It seems Python wants to print a nice traceback message for KeyboardInterrupt, but the spawned thread constantly steals the lock for stdout, so Python just breaks.

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