On Fri, 4 Jan 2008, Matt Mackall wrote:

quoted text
> > 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.

quoted text
> 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.

quoted text
> 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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