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slub: Do not use 192 byte sized cache if minimum alignment is 128 byte

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The 192 byte cache is not necessary if we have a basic alignment of 128
byte. If it would be used then the 192 would be aligned to the next 128 byte
boundary which would result in another 256 byte cache. Two 256 kmalloc caches
cause sysfs to complain about a duplicate entry.

MIPS needs 128 byte aligned kmalloc caches and spits out warnings on boot without
this patch.

Signed-off-by: Christoph Lameter <cl@linux-foundation.org>
Signed-off-by: Pekka Enberg <penberg@cs.helsinki.fi>
This commit is contained in:
Christoph Lameter 2008-07-03 09:14:26 -05:00 committed by Pekka Enberg
parent 481c5346d0
commit 41d54d3bf8
2 changed files with 12 additions and 2 deletions
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2
include/linux/slub_def.h
View File

@ -137,10 +137,12 @@ static __always_inline int kmalloc_index(size_t size)
if (size <= KMALLOC_MIN_SIZE)
return KMALLOC_SHIFT_LOW;
#if KMALLOC_MIN_SIZE <= 64
if (size > 64 && size <= 96)
return 1;
if (size > 128 && size <= 192)
return 2;
#endif
if (size <= 8) return 3;
if (size <= 16) return 4;
if (size <= 32) return 5;

12
mm/slub.c
View File

@ -2995,8 +2995,6 @@ void __init kmem_cache_init(void)
create_kmalloc_cache(&kmalloc_caches[1],
"kmalloc-96", 96, GFP_KERNEL);
caches++;
}
if (KMALLOC_MIN_SIZE <= 128) {
create_kmalloc_cache(&kmalloc_caches[2],
"kmalloc-192", 192, GFP_KERNEL);
caches++;
@ -3026,6 +3024,16 @@ void __init kmem_cache_init(void)
for (i = 8; i < KMALLOC_MIN_SIZE; i += 8)
size_index[(i - 1) / 8] = KMALLOC_SHIFT_LOW;
if (KMALLOC_MIN_SIZE == 128) {
/*
* The 192 byte sized cache is not used if the alignment
* is 128 byte. Redirect kmalloc to use the 256 byte cache
* instead.
*/
for (i = 128 + 8; i <= 192; i += 8)
size_index[(i - 1) / 8] = 8;
}
slab_state = UP;
/* Provide the correct kmalloc names now that the caches are up */