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mem_slab: enforce minimum alignment on statically allocated slabs

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There is no point allowing smaller alignments. And on 64-bit systems the
minimum becomes 8 rather than 4, so let's adjust things automatically.

Signed-off-by: Nicolas Pitre <npitre@baylibre.com>
This commit is contained in:
Nicolas Pitre 2019-05-21 21:40:38 -04:00 • committed by Anas Nashif
commit 46cd5a0330
1 changed files with 5 additions and 4 deletions
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9
include/kernel.h
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@ -3912,11 +3912,11 @@ struct k_mem_slab {
* @req K-MSLAB-001
*/
#define K_MEM_SLAB_DEFINE(name, slab_block_size, slab_num_blocks, slab_align) \
char __noinit __aligned(slab_align) \
_k_mem_slab_buf_##name[(slab_num_blocks) * (slab_block_size)]; \
char __noinit __aligned(WB_UP(slab_align)) \
_k_mem_slab_buf_##name[(slab_num_blocks) * WB_UP(slab_block_size)]; \
Z_STRUCT_SECTION_ITERABLE(k_mem_slab, name) = \
_K_MEM_SLAB_INITIALIZER(name, _k_mem_slab_buf_##name, \
slab_block_size, slab_num_blocks)
WB_UP(slab_block_size), slab_num_blocks)
/**
* @brief Initialize a memory slab.
@ -3925,7 +3925,8 @@ struct k_mem_slab {
*
* The memory slab's buffer contains @a slab_num_blocks memory blocks
* that are @a slab_block_size bytes long. The buffer must be aligned to an
* N-byte boundary, where N is a power of 2 larger than 2 (i.e. 4, 8, 16, ...).
* N-byte boundary matching a word boundary, where N is a power of 2
* (i.e. 4 on 32-bit systems, 8, 16, ...).
* To ensure that each memory block is similarly aligned to this boundary,
* @a slab_block_size must also be a multiple of N.
*