Re: Slub Allocator: Why get_order(size * MAX_OBJS_PER_PAGE) - 1 in function slab_order()?

[David Rientjes <rientjes@google.com>]

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.