Make CPython segfault in 5 lines of code

Re:Make CPython segfault in 5 lines of code by paulddraper on Wednesday December 18, 2019 @11:25PM (#21831998)

If # of lines are important, this problem can actually be demonstrated in 1 line:

    for x, x.__new__ in [(__import__('queue').Full, print)]: __import__('glob').iglob(0).throw(x)

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Re:Make CPython segfault in 5 lines of code by filmor on Thursday December 19, 2019 @01:42AM (#21832562)

This doesn't trigger a segfault for me in Python 3.7, just an exception that `print_exception` expects an `Exception`.

Someone commented below the gist with this one-liner:

    (i for i in []).throw(type('E', (BaseException,), dict(__new__=lambda cls, *args: cls))())

I managed to golf it a bit down to this ;):

    n="__new__";(i for i in []).throw(type(n,(IOError,),{n:lambda c,*a:c})())

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Re:Make CPython segfault in 5 lines of code by ehsankia on Thursday December 19, 2019 @01:51AM (#21832622)

Just one character but replace `[]` with `n` too :)

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Re:Make CPython segfault in 5 lines of code by chrismorgan on Thursday December 19, 2019 @07:13AM (#21834192)

Or save that character by removing the space before `[]` instead (which you can’t do if you write `n`).

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Re:Make CPython segfault in 5 lines of code by paulddraper on Thursday December 19, 2019 @07:26AM (#21834268)

Mine produces segfault with

    $ python3.7 -VV
    Python 3.7.5 (default, Nov  7 2019, 10:50:52) 
    [GCC 8.3.0]

Some more golfing with yours:

    (i for i in[]).throw(type('',(IOError,),{'__new__':lambda a,*b:a}))

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