Messages in this thread | | | Date | Sat, 5 Jan 2008 12:16:22 -0800 (PST) | From | Christoph Lameter <> | Subject | Re: [PATCH] procfs: provide slub's /proc/slabinfo |
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On Fri, 4 Jan 2008, Matt Mackall wrote:
> > SLUB 32 (all memory of the 4k page is used for 128 byte objects) > > SLAB 29/30 (management structure occupies first two/three objects) > > SLOB 30(?) (Alignment results in object being 136 byte of effective size, > > we have 16 bytes leftover that could be used for a > > very small allocation. Right?) > > Don't know how you got to 136, the minimum alignment is 4 on x86. But I
Right I am thinking about 64 bit systems where the alignment is 8 bytes.
> already said in my last email that SLUB would win for the special case > of power of two allocations. But as long as we're looking at worst > cases, let's consider an alloc of 257 bytes..
Yup that hits it by forcing a rounding up to a size of 512 bytes because there is no intermediate cache size before 1024. The rounding up is a pretty weak spot in terms of memory use.
> SLUB 8 (1016 bytes wasted) > SLOB 15 (105 bytes wasted, with 136 bytes still usable)
Well we can actually turn this around. What I gave was not actually the worst case for SLOB. The worst case is an 8 byte allocation where SLOB needs double the memory of SLUB.
SLUB 512 (Nothing wasted) SLOB 256 (Half of the page wasted for metadata) SLAB 119 (32 byte mininum alloc size + management struct needs)
But these are all extreme cases. Depends on the mix of allocs who wins and from what I can tell the avoiding of rounding up to a power of two gives SLOB a key advantage. If we would find the worst offenders there and use kmem_cache_alloc instead then we may be able to offset that advantage.
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