92 lines
4.0 KiB
Plaintext
92 lines
4.0 KiB
Plaintext
This document describes some caveats about the use of Valgrind with
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Python. Valgrind is used periodically by Python developers to try
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to ensure there are no memory leaks or invalid memory reads/writes.
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If you don't want to read about the details of using Valgrind, there
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are still two things you must do to suppress the warnings. First,
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you must use a suppressions file. One is supplied in
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Misc/valgrind-python.supp. Second, you must do one of the following:
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* Uncomment Py_USING_MEMORY_DEBUGGER in Objects/obmalloc.c,
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then rebuild Python
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* Uncomment the lines in Misc/valgrind-python.supp that
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suppress the warnings for PyObject_Free and PyObject_Realloc
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If you want to use Valgrind more effectively and catch even more
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memory leaks, you will need to configure python --without-pymalloc.
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PyMalloc allocates a few blocks in big chunks and most object
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allocations don't call malloc, they use chunks doled about by PyMalloc
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from the big blocks. This means Valgrind can't detect
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many allocations (and frees), except for those that are forwarded
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to the system malloc. Note: configuring python --without-pymalloc
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makes Python run much slower, especially when running under Valgrind.
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You may need to run the tests in batches under Valgrind to keep
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the memory usage down to allow the tests to complete. It seems to take
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about 5 times longer to run --without-pymalloc.
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Details:
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--------
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Python uses its own small-object allocation scheme on top of malloc,
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called PyMalloc.
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Valgrind may show some unexpected results when PyMalloc is used.
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Starting with Python 2.3, PyMalloc is used by default. You can disable
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PyMalloc when configuring python by adding the --without-pymalloc option.
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If you disable PyMalloc, most of the information in this document and
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the supplied suppressions file will not be useful. As discussed above,
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disabling PyMalloc can catch more problems.
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If you use valgrind on a default build of Python, you will see
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many errors like:
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==6399== Use of uninitialised value of size 4
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==6399== at 0x4A9BDE7E: PyObject_Free (obmalloc.c:711)
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==6399== by 0x4A9B8198: dictresize (dictobject.c:477)
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These are expected and not a problem. Tim Peters explains
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the situation:
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PyMalloc needs to know whether an arbitrary address is one
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that's managed by it, or is managed by the system malloc.
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The current scheme allows this to be determined in constant
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time, regardless of how many memory areas are under pymalloc's
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control.
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The memory pymalloc manages itself is in one or more "arenas",
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each a large contiguous memory area obtained from malloc.
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The base address of each arena is saved by pymalloc
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in a vector. Each arena is carved into "pools", and a field at
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the start of each pool contains the index of that pool's arena's
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base address in that vector.
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Given an arbitrary address, pymalloc computes the pool base
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address corresponding to it, then looks at "the index" stored
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near there. If the index read up is out of bounds for the
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vector of arena base addresses pymalloc maintains, then
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pymalloc knows for certain that this address is not under
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pymalloc's control. Otherwise the index is in bounds, and
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pymalloc compares
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the arena base address stored at that index in the vector
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to
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the arbitrary address pymalloc is investigating
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pymalloc controls this arbitrary address if and only if it lies
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in the arena the address's pool's index claims it lies in.
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It doesn't matter whether the memory pymalloc reads up ("the
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index") is initialized. If it's not initialized, then
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whatever trash gets read up will lead pymalloc to conclude
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(correctly) that the address isn't controlled by it, either
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because the index is out of bounds, or the index is in bounds
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but the arena it represents doesn't contain the address.
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This determination has to be made on every call to one of
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pymalloc's free/realloc entry points, so its speed is critical
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(Python allocates and frees dynamic memory at a ferocious rate
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-- everything in Python, from integers to "stack frames",
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lives in the heap).
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