linux-kernel - Slub Allocator: Why get_order(size * MAX_OBJS_PER_PAGE)

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Message-ID: <CABexPfEa-kTdH3=7pa=geVE5rUSiuj2Gzt6Zbjz_-cwXJVYGFA@mail.gmail.com>
Date:	Thu, 24 Nov 2011 12:44:04 +0800
From:	zhihua che <zhihua.che@...il.com>
To:	linux-kernel@...r.kernel.org
Subject: Slub Allocator: Why get_order(size * MAX_OBJS_PER_PAGE) - 1 in
 function slab_order()?

2011/11/24 David Rientjes <rientjes@...gle.com>:
> On Wed, 23 Nov 2011, zhihua che wrote:
>
>> I know what you mean, that is, a slab can only store no more than
>> MAX_OBJS_PER_PAGE, actually 0x7FFF, objects.
>>
>> But  get_order(size * MAX_OBJS_PER_PAGE) already returns the order
>> which reserves no_more_than size * MAX_OBJS_PER_PAGE memory.  Right?
>>
>
> Yes, but it reserves too much memory if the conditional is true.
>
>> So I think there is no need to subtract one.
>>
>
> If we didn't subtract one, then the order of a slab page would allow for
> _more_ than MAX_OBJS_PER_PAGE to be allocated and that's not allowed
> because of the restrictions in struct page.
>
> Consider a page size of 4K and an object size of 8 bytes.
> get_order(8 * 32767) would be 6, so that's a 4K * 2^6 = 256K slab page
> without the subtraction and could allocate (256K * 1024 / 8) = 32768 which
> is greater than MAX_OBJS_PER_PAGE and not allowed.
>
> So we subtract one so the compound slab page is guaranteed to allocate
> less than MAX_OBJS_PER_PAGE.

Oh, sorry to bother you ... I guess I calculated a wrong result. I
repeat your and my example carefully and you're right.

Thanks very much for your patience :)
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