Issue1110055
Created on 2005-01-26 17:18 by jengeldk, last changed 2022-04-11 14:56 by admin. This issue is now closed.
| Messages (4) | |||
|---|---|---|---|
| msg24058 - (view) | Author: Jacob Engelbrecht (jengeldk) | Date: 2005-01-26 17:18 | |
when running recursive function you get a coredump with
deep recursion.
eg
from sys import *
n = 30000
setrecursionlimit(n+1)
def fact(n):
if n==1:
return 1
return fact(n-1)*n
fact(n)
This is seen on linux i686 with both python2.3 and
python2.4, the recursion depth which triggers the core
dump is 26211 with python2.4 and 29123 with python2.3
with a machine having 2076860 kB of memory, on machines
with less memory smaller numbers are seen.
this is what gdb tells me:
jacob@pauling:/scratch/jacob$ gdb /usr/bin/python2.4 core
GNU gdb 6.3-debian
Copyright 2004 Free Software Foundation, Inc.
GDB is free software, covered by the GNU General Public
License, and you are
welcome to change it and/or distribute copies of it
under certain conditions.
Type "show copying" to see the conditions.
There is absolutely no warranty for GDB. Type "show
warranty" for details.
This GDB was configured as "i386-linux"...(no debugging
symbols found)
Using host libthread_db library
"/lib/tls/libthread_db.so.1".
(no debugging symbols found)
Core was generated by `python2.4 /home/user_4/jacob/rr
30000'.
Program terminated with signal 11, Segmentation fault.
warning: current_sos: Can't read pathname for load map:
Input/output error
Reading symbols from /lib/tls/libpthread.so.0...(no
debugging symbols found)...done.
Loaded symbols for /lib/tls/libpthread.so.0
Reading symbols from /lib/tls/libdl.so.2...(no
debugging symbols found)...done.
Loaded symbols for /lib/tls/libdl.so.2
Reading symbols from /lib/tls/libutil.so.1...(no
debugging symbols found)...done.
Loaded symbols for /lib/tls/libutil.so.1
Reading symbols from /lib/tls/libm.so.6...
(no debugging symbols found)...done.
Loaded symbols for /lib/tls/libm.so.6
Reading symbols from /lib/tls/libc.so.6...(no debugging
symbols found)...done.
Loaded symbols for /lib/tls/libc.so.6
Reading symbols from /lib/ld-linux.so.2...(no debugging
symbols found)...done.
Loaded symbols for /lib/ld-linux.so.2
#0 0x400c94bf in mallopt () from /lib/tls/libc.so.6
|
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| msg24059 - (view) | Author: Grant Olson (logistix) | Date: 2005-01-29 20:12 | |
Logged In: YES
user_id=699438
This looks like a stack overflow. There's not much python
can do when the stack runs out of memory, which is why the
default recursion limit is set to 1000.
Also, at least in the reproducable, the number you are building
is going to consume excessive amounts of memory.
I ran this test (to avoid creating a giant long number) and still
got the segfault.
from sys import *
n = 30000
setrecursionlimit(n+1)
def nofact(n):
if n==1:
return 1
return nofact(n-1)
nofact(n)
|
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| msg24060 - (view) | Author: Josiah Carlson (josiahcarlson) * ![]() |
Date: 2005-01-29 20:16 | |
Logged In: YES user_id=341410 The fact that an error occurs is not surprising. Python is limited by the C stack size, which from what I understand is well below 2GB. The fact that it gets to nearly 30k levels of recursion for you is amazing to me, the most used platform (Windows) can only ever get to around 5500 levels before they get "MemoryError: stack overflow" exceptions. I believe that the reason you are getting a segfault on linux is the way the linux malloc() and free() work. Specifically, malloc() on linux will give you a pointer to memory, regardless of whether it is available. If your program has used up all of its stack space, and you need more, the pointer will be invalid. If Python happens to call a free() before you actually access the invalid pointer, everything will work. If Python doesn't call free() before you access the invalid pointer, you get a segfault. Unfortunately, there is no way that Python (or any other program on linux) can know that the pointer it has gotten from malloc() is invalid. Furthermore, as per the docs here: http://docs.python.org/lib/module-sys.html " setrecursionlimit(limit) Set the maximum depth of the Python interpreter stack to limit. This limit prevents infinite recursion from causing an overflow of the C stack and crashing Python. The highest possible limit is platform-dependent. A user may need to set the limit higher when she has a program that requires deep recursion and a platform that supports a higher limit. This should be done with care, because a too-high limit can lead to a crash." I would suggest that this bug be closed as "3rd party, will not fix", and suggest the OP read the documentation. |
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| msg24061 - (view) | Author: Tim Peters (tim.peters) * ![]() |
Date: 2005-01-29 20:20 | |
Logged In: YES user_id=31435 The docs for setrecursionlimit() are indeed already quite clear that you boost its default value at your own risk. So this isn't a bug, it's a documented limitation. |
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| History | |||
|---|---|---|---|
| Date | User | Action | Args |
| 2022-04-11 14:56:09 | admin | set | github: 41493 |
| 2005-01-26 17:18:09 | jengeldk | create | |

