python segfault

Michael Poeltl michael.poeltl at univie.ac.at
Wed Mar 28 08:16:50 CEST 2012


hi,

* Dave Angel <d at davea.name> [2012-03-28 04:38]:
> On 03/27/2012 06:27 PM, Michael Poeltl wrote:
> >hi,
> >
> >can anybody tell why this 'little stupid *thing* of code' let's python-3.2.2, 2.6.X or python 2.7.2 segfault?
> >
> >>>def get_steps2(pos=0, steps=0):
> >...     if steps == 0:
> >...         pos = random.randint(-1,1)
> >...     if pos == 0:
> >...         return steps
> >...     steps += 2
> >...     pos += random.randint(-1,1)
> >...     return get_steps2(pos,steps)
> >...
> ><SNIP>
> >0
> >2
> >8
> >0
> >Segmentation fault
> >?>
> >
> >funny, isn't it?
> >I was able to reproduce this segfault on various machines (32bit 64bit), ubuntu, slackware, debian
> >python.X segfaults on all of them
> >
> >thx
> >Michael
> 
> Others have explained why you can't just raise the recursion limit
> to arbitrarily large values, and why there's no particular bound on
> the possible recursion size.  But the real question is why you don't
> do the completely trivial conversion to a non-recursive equivalent.
> 
> All you need do is add a while True:  to the beginning of the
> function, and remove the return statement.
yeah - of course 'while True' was the first, most obvious best way... ;-)
but I was asked if there was a way without 'while True'
and so I started the 'recursive function'

and quick quick; RuntimeError-Exception -> not thinking much -> just adding
two zeros to the default limit (quick and dirty) -> segfault ==> subject: python segfault ;-)

and that was my first time that I received a segfault and not an Exception

NOW it's quite clear ;-)

thank you!
Michael
> 
> 
> 
> -- 
> 
> DaveA
> 

-- 
Michael Poeltl
Computational Materials Physics      voice: +43-1-4277-51409
Univ. Wien, Sensengasse 8/12         fax:   +43-1-4277-9514 (or 9513) 
A-1090 Wien, AUSTRIA   cmp.mpi.univie.ac.at 
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