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| 1 | +This document describes some caveats about the use of Valgrind with |
| 2 | +Python. Valgrind is used periodically by Python developers to try |
| 3 | +to ensure there are no memory leaks or invalid memory reads/writes. |
| 4 | + |
| 5 | +If you don't want to read about the details of using Valgrind, there |
| 6 | +are still two things you must do to suppress the warnings. First, |
| 7 | +you must use a suppressions file. One is supplied in |
| 8 | +Misc/valgrind-python.supp. Second, you must do one of the following: |
| 9 | + |
| 10 | + * Uncomment Py_USING_MEMORY_DEBUGGER in Objects/obmalloc.c, |
| 11 | + then rebuild Python |
| 12 | + * Uncomment the lines in Misc/valgrind-python.supp that |
| 13 | + suppress the warnings for PyObject_Free and PyObject_Realloc |
| 14 | + |
| 15 | +Details: |
| 16 | +-------- |
| 17 | +Python uses its own allocation scheme on top of malloc called PyMalloc. |
| 18 | +Valgrind my show some unexpected results when PyMalloc is used. |
| 19 | +Starting with Python 2.3, PyMalloc is used by default. You can disable |
| 20 | +PyMalloc when configuring python by adding the --without-pymalloc option. |
| 21 | +If you disable PyMalloc, most of the information in this document and |
| 22 | +the supplied suppressions file will not be useful. |
| 23 | + |
| 24 | +If you use valgrind on a default build of Python, you will see |
| 25 | +many errors like: |
| 26 | + |
| 27 | + ==6399== Use of uninitialised value of size 4 |
| 28 | + ==6399== at 0x4A9BDE7E: PyObject_Free (obmalloc.c:711) |
| 29 | + ==6399== by 0x4A9B8198: dictresize (dictobject.c:477) |
| 30 | + |
| 31 | +These are expected and not a problem. Tim Peters explains |
| 32 | +the situation: |
| 33 | + |
| 34 | + PyMalloc needs to know whether an arbitrary address is one |
| 35 | + that's managed by it, or is managed by the system malloc. |
| 36 | + The current scheme allows this to be determined in constant |
| 37 | + time, regardless of how many memory areas are under pymalloc's |
| 38 | + control. |
| 39 | + |
| 40 | + The memory pymalloc manages itself is in one or more "arenas", |
| 41 | + each a large contiguous memory area obtained from malloc. |
| 42 | + The base address of each arena is saved by pymalloc |
| 43 | + in a vector, and a field at the start of each arena contains |
| 44 | + the index of that arena's base address in that vector. |
| 45 | + |
| 46 | + Given an arbitrary address, pymalloc computes the arena base |
| 47 | + address corresponding to it, then looks at "the index" stored |
| 48 | + near there. If the index read up is out of bounds for the |
| 49 | + vector of arena base addresses pymalloc maintains, then |
| 50 | + pymalloc knows for certain that this address is not under |
| 51 | + pymalloc's control. Otherwise the index is in bounds, and |
| 52 | + pymalloc compares |
| 53 | + |
| 54 | + the arena base address stored at that index in the vector |
| 55 | + |
| 56 | + to |
| 57 | + |
| 58 | + the computed arena address |
| 59 | + |
| 60 | + pymalloc controls this arena if and only if they're equal. |
| 61 | + |
| 62 | + It doesn't matter whether the memory pymalloc reads up ("the |
| 63 | + index") is initialized. If it's not initialized, then |
| 64 | + whatever trash gets read up will lead pymalloc to conclude |
| 65 | + (correctly) that the address isn't controlled by it. |
| 66 | + |
| 67 | + This determination has to be made on every call to one of |
| 68 | + pymalloc's free/realloc entry points, so its speed is critical |
| 69 | + (Python allocates and frees dynamic memory at a ferocious rate |
| 70 | + -- everything in Python, from integers to "stack frames", |
| 71 | + lives in the heap). |
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