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Fixed size calculation in MALLOC memory pool creation macro.

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The LWIP_MALLOC_MEMPOOL macro needs to use the aligned size of the
memp_malloc_helper structure, since mem_alloc() uses it to calculate
the required pool element size. If LWIP_MEM_ALIGN_SIZE(x) is redefined
to align to something larger than 4, then in some cases
the current code can lead to unexpected mem_alloc() failures.
For example:
    #define LWIP_MEM_ALIGN_SIZE(size)       (((size) + 31) & ~31)
and the largest MALLOC pool is of size 60 bytes, e.g.:
    #define LWIP_MALLOC_MEMPOOL(256, 60)
then the following call:
    mem_malloc(58)
will cause an assertion.
This commit is contained in:
Stathis Voukelatos 2013-01-17 16:52:47 +00:00 committed by Simon Goldschmidt
parent a70567e74f
commit 38bfe50508
1 changed files with 1 additions and 1 deletions
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2
src/include/lwip/memp_std.h
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@ -12,7 +12,7 @@
#ifndef LWIP_MALLOC_MEMPOOL
/* This treats "malloc pools" just like any other pool.
The pools are a little bigger to provide 'size' as the amount of user data. */
#define LWIP_MALLOC_MEMPOOL(num, size) LWIP_MEMPOOL(POOL_##size, num, (size + sizeof(struct memp_malloc_helper)), "MALLOC_"#size)
#define LWIP_MALLOC_MEMPOOL(num, size) LWIP_MEMPOOL(POOL_##size, num, (size + LWIP_MEM_ALIGN_SIZE(sizeof(struct memp_malloc_helper))), "MALLOC_"#size)
#define LWIP_MALLOC_MEMPOOL_START
#define LWIP_MALLOC_MEMPOOL_END
#endif /* LWIP_MALLOC_MEMPOOL */