OXIESEC PANEL

Name	Size	Modified	Perms	Actions
📁 ..	-	04/13/2024 08:55:16 PM	rwxr-xr-x
📄 README.valgrind	4.67 KB	06/03/2025 06:47:52 PM	rw-r--r--	Delete
📄 gdbinit	4.85 KB	06/03/2025 06:47:52 PM	rw-r--r--	Delete
📄 valgrind-python.supp	8.62 KB	06/03/2025 06:47:52 PM	rw-r--r--	Delete

Editing: README.valgrind

This document describes some caveats about the use of Valgrind with
Python.  Valgrind is used periodically by Python developers to try
to ensure there are no memory leaks or invalid memory reads/writes.

If you want to enable valgrind support in Python, you will need to
configure Python --with-valgrind option or an older option
--without-pymalloc.

UPDATE: Python 3.6 now supports PYTHONMALLOC=malloc environment variable which
can be used to force the usage of the malloc() allocator of the C library.

If you don't want to read about the details of using Valgrind, there
are still two things you must do to suppress the warnings.  First,
you must use a suppressions file.  One is supplied in
Misc/valgrind-python.supp.  Second, you must uncomment the lines in 
Misc/valgrind-python.supp that suppress the warnings for PyObject_Free and
PyObject_Realloc.

If you want to use Valgrind more effectively and catch even more
memory leaks, you will need to configure python --without-pymalloc.
PyMalloc allocates a few blocks in big chunks and most object
allocations don't call malloc, they use chunks doled about by PyMalloc
from the big blocks.  This means Valgrind can't detect
many allocations (and frees), except for those that are forwarded
to the system malloc.  Note: configuring python --without-pymalloc
makes Python run much slower, especially when running under Valgrind.
You may need to run the tests in batches under Valgrind to keep
the memory usage down to allow the tests to complete.  It seems to take
about 5 times longer to run --without-pymalloc.

Apr 15, 2006:
  test_ctypes causes Valgrind 3.1.1 to fail (crash).
  test_socket_ssl should be skipped when running valgrind.
	The reason is that it purposely uses uninitialized memory.
	This causes many spurious warnings, so it's easier to just skip it.

Details:
--------
Python uses its own small-object allocation scheme on top of malloc,
called PyMalloc.

Valgrind may show some unexpected results when PyMalloc is used.
Starting with Python 2.3, PyMalloc is used by default.  You can disable
PyMalloc when configuring python by adding the --without-pymalloc option.
If you disable PyMalloc, most of the information in this document and
the supplied suppressions file will not be useful.  As discussed above,
disabling PyMalloc can catch more problems.

PyMalloc uses 256KB chunks of memory, so it can't detect anything
wrong within these blocks.  For that reason, compiling Python
--without-pymalloc usually increases the usefulness of other tools.

If you use valgrind on a default build of Python,  you will see
many errors like:

==6399== Use of uninitialised value of size 4
        ==6399== at 0x4A9BDE7E: PyObject_Free (obmalloc.c:711)
        ==6399== by 0x4A9B8198: dictresize (dictobject.c:477)

These are expected and not a problem.  Tim Peters explains
the situation:

PyMalloc needs to know whether an arbitrary address is one
	that's managed by it, or is managed by the system malloc.
	The current scheme allows this to be determined in constant
	time, regardless of how many memory areas are under pymalloc's
	control.

The memory pymalloc manages itself is in one or more "arenas",
	each a large contiguous memory area obtained from malloc.
	The base address of each arena is saved by pymalloc
	in a vector.  Each arena is carved into "pools", and a field at
	the start of each pool contains the index of that pool's arena's
	base address in that vector.

Given an arbitrary address, pymalloc computes the pool base
	address corresponding to it, then looks at "the index" stored
	near there.  If the index read up is out of bounds for the
	vector of arena base addresses pymalloc maintains, then
	pymalloc knows for certain that this address is not under
	pymalloc's control.  Otherwise the index is in bounds, and
	pymalloc compares

the arena base address stored at that index in the vector

the arbitrary address pymalloc is investigating

pymalloc controls this arbitrary address if and only if it lies
        in the arena the address's pool's index claims it lies in.

It doesn't matter whether the memory pymalloc reads up ("the
	index") is initialized.  If it's not initialized, then
	whatever trash gets read up will lead pymalloc to conclude
	(correctly) that the address isn't controlled by it, either
	because the index is out of bounds, or the index is in bounds
	but the arena it represents doesn't contain the address.

This determination has to be made on every call to one of
	pymalloc's free/realloc entry points, so its speed is critical
	(Python allocates and frees dynamic memory at a ferocious rate
	-- everything in Python, from integers to "stack frames",
	lives in the heap).