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SubjectRe: [RFC 0/2] guarantee natural alignment for kmalloc()
On Wed, Mar 20, 2019 at 10:48:03PM +0100, Vlastimil Babka wrote:
> On 3/20/2019 7:53 PM, Matthew Wilcox wrote:
> > On Wed, Mar 20, 2019 at 09:48:47AM +0100, Vlastimil Babka wrote:
> >> Natural alignment to size is rather well defined, no? Would anyone ever
> >> assume a larger one, for what reason?
> >> It's now where some make assumptions (even unknowingly) for natural
> >> There are two 'odd' sizes 96 and 192, which will keep cacheline size
> >> alignment, would anyone really expect more than 64 bytes?
> >
> > Presumably 96 will keep being aligned to 32 bytes, as aligning 96 to 64
> > just results in 128-byte allocations.
>
> Well, looks like that's what happens. This is with SLAB, but the alignment
> calculations should be common:
>
> slabinfo - version: 2.1
> # name <active_objs> <num_objs> <objsize> <objperslab> <pagesperslab> : tunables <limit> <batchcount> <sharedfactor> : slabdata <active_slabs> <num_slabs> <sharedavail>
> kmalloc-96 2611 4896 128 32 1 : tunables 120 60 8 : slabdata 153 153 0
> kmalloc-128 4798 5536 128 32 1 : tunables 120 60 8 : slabdata 173 173 0

Hmm. On my laptop, I see:

kmalloc-96 28050 35364 96 42 1 : tunables 0 0 0 : slabdata 842 842 0

That'd take me from 842 * 4k pages to 1105 4k pages -- an extra megabyte of
memory.

This is running Debian's 4.19 kernel:

# CONFIG_SLAB is not set
CONFIG_SLUB=y
# CONFIG_SLOB is not set
CONFIG_SLAB_MERGE_DEFAULT=y
CONFIG_SLAB_FREELIST_RANDOM=y
CONFIG_SLAB_FREELIST_HARDENED=y
CONFIG_SLUB_CPU_PARTIAL=y


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