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SubjectRe: Slub Allocator: Why get_order(size * MAX_OBJS_PER_PAGE) - 1 in function slab_order()?
On Wed, 23 Nov 2011, zhihua che wrote:

> Hi, everyone,
> I'm reading the kernel codes about slub allocator and I come
> across a confusion. Precisely, I'm reading the initialization of the
> slub allocator, kmem_cache_init(), and I find it needs call
> calculate_sizes() to determine the order of a kmem_cache, given the
> size of the object. In turn, it calls the get_order() to get a
> possible order. The problem is, in the start of this function, why it
> looks like this:
>
> if (order_objects(min_order, size, reserved) > MAX_OBJS_PER_PAGE)
> return get_order(size * MAX_OBJS_PER_PAGE) - 1;
>
> I don't know why it subtracts one from the order returned by
> get_order().
> because as far as I know, get_order() returns the order the
> slab requires to reserve size * MAX_OBJS_PER_PAGE memory. If it
> subtracts 1 from the order returned by get_order(), the slab can't
> store MAX_OBJS_PER_PAGE objects at all, instead it can only store half
> of the MAX_OBJS_PER_PAGE objects.
> Could you correct me if I think in a wrong way.

I agree it looks confusing, but it's correct. SLUB can only store
MAX_OBJS_PER_PAGE because of limitations in struct page (see the comments
in include/linux/mm_types.h). So if the order will yield a page that
could fit _more_ than MAX_OBJS_PER_PAGE, we need to reduce the order by a
factor of 1.


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