linux-kernel - Re: [RFC] slub: ARCH_SLAB_MINALIGN defaults to 8 on x86

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Date:	Wed, 03 Feb 2010 17:25:03 +0000
From:	Richard Kennedy <richard@....demon.co.uk>
To:	Christoph Lameter <cl@...ux-foundation.org>
Cc:	penberg <penberg@...helsinki.fi>, Ingo Molnar <mingo@...e.hu>,
	Thomas Gleixner <tglx@...utronix.de>,
	lkml <linux-kernel@...r.kernel.org>,
	linux-mm <linux-mm@...ck.org>
Subject: Re: [RFC] slub: ARCH_SLAB_MINALIGN defaults to 8 on x86_32. is
 this too big?

On Wed, 2010-02-03 at 09:41 -0600, Christoph Lameter wrote:
> On Wed, 3 Feb 2010, Richard Kennedy wrote:
> 
> > slub.c sets the default value of ARCH_SLAB_MINALIGN to sizeof(unsigned
> > long long) if the architecture didn't already override it.
> >
> > And as x86_32 doesn't set a value this means that slab objects get
> > aligned to 8 bytes, potentially wasting 4 bytes per object. Slub forces
> > objects to be aligned to sizeof(void *) anyway, but I don't see that
> > there is any need for it to be 8 on 32bits.
> 
> Note that 64 bit entities may  exist even under 32 bit (llong) that need
> to be aligned properly.
> 
> struct buffer_head contains a sector_t which is 64 bit so you should align
> to an 8 byte boundary.
> 
> > I'm working on a patch to pack more buffer_heads into each kmem_cache
> > slab page.
> > On 32 bits the structure size is 52 bytes and with the alignment applied
> > I end up with a slab of 73 x 56 byte objects. However, if the minimum
> > alignment was sizeof(void *) then I'd get 78 x 52 byte objects. So there
> > is quite a memory saving to be had in changing this.
> 
> SLUB is not restricted to order 0 pages and can use order 1 or 2 pages as
> long as this reduces the memory footprint (byte wastage in a slab page is
> reduced) and as long as the kernel has contiguous memory available. It
> will use order 0 when memory is fragmented.
> 
> > Can I define a ARCH_SLAB_MINALIGN in x86_64 to sizeof(void *) ?
> > or would it be ok to change the default in slub.c to sizeof(void *) ?
> >
> > Or am I missing something ?
> 
> I'd say leave it alone.

I definitely don't want to break the alignment ;) but gcc aligns
unsigned long long on 4 byte boundaries on 32 bit.

Running this test code :-

#ifdef __compiler_offsetof
#define offsetof(TYPE,MEMBER) __compiler_offsetof(TYPE,MEMBER)
#else
#define offsetof(TYPE, MEMBER) ((size_t) &((TYPE *)0)->MEMBER)
#endif

struct test_align {
	char c;
	unsigned long long l;
};
void main() {
printf( "size = %d , offset of l = $d\n", 
	sizeof(struct test_align),
	offsetof(struct test_align,l) );
}

gives me this output :- 
32 bit : size = 12 , offset of l = 4
64 bit : size = 16 , offset of l = 8

Doesn't that suggest that it would be safe to use sizeof(void *) ?
(at least on x86 anyway).

We end up with a large number of buffer_heads and as they are pretty
small an extra 4 bytes does make a significant difference. 
On my 64 bit machine I often see thousands of pages of buffer_heads, so
squeezing a few more per page could be a considerable saving.

regards
Richard





 


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