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Subject: Re: [RFC 0/2] guarantee natural alignment for kmalloc()
To:     Matthew Wilcox <willy@infradead.org>
Cc:     Christopher Lameter <cl@linux.com>, linux-mm@kvack.org,
        Pekka Enberg <penberg@kernel.org>,
        David Rientjes <rientjes@google.com>,
        Joonsoo Kim <iamjoonsoo.kim@lge.com>,
        Ming Lei <ming.lei@redhat.com>,
        Dave Chinner <david@fromorbit.com>,
        "Darrick J . Wong" <darrick.wong@oracle.com>,
        Christoph Hellwig <hch@lst.de>,
        Michal Hocko <mhocko@kernel.org>, linux-kernel@vger.kernel.org,
        linux-xfs@vger.kernel.org, linux-fsdevel@vger.kernel.org,
        linux-block@vger.kernel.org
References: <20190319211108.15495-1-vbabka@suse.cz>
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 <5d7fee9c-1a80-6ac9-ac1d-b1ce05ed27a8@suse.cz>
 <20190320185347.GZ19508@bombadil.infradead.org>
From:   Vlastimil Babka <vbabka@suse.cz>
Message-ID: <b5290e04-6f29-c237-78a7-511821183efe@suse.cz>
Date:   Wed, 20 Mar 2019 22:48:03 +0100
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On 3/20/2019 7:53 PM, Matthew Wilcox wrote:
> On Wed, Mar 20, 2019 at 09:48:47AM +0100, Vlastimil Babka wrote:
>> Natural alignment to size is rather well defined, no? Would anyone ever
>> assume a larger one, for what reason?
>> It's now where some make assumptions (even unknowingly) for natural
>> There are two 'odd' sizes 96 and 192, which will keep cacheline size
>> alignment, would anyone really expect more than 64 bytes?
> 
> Presumably 96 will keep being aligned to 32 bytes, as aligning 96 to 64
> just results in 128-byte allocations.

Well, looks like that's what happens. This is with SLAB, but the alignment
calculations should be common: 

slabinfo - version: 2.1
# name            <active_objs> <num_objs> <objsize> <objperslab> <pagesperslab> : tunables <limit> <batchcount> <sharedfactor> : slabdata <active_slabs> <num_slabs> <sharedavail>
kmalloc-96          2611   4896    128   32    1 : tunables  120   60    8 : slabdata    153    153      0
kmalloc-128         4798   5536    128   32    1 : tunables  120   60    8 : slabdata    173    173      0