Re: [Question] qemu-img convert block alignment

[Zhenyu Ye <yezhenyu2@huawei.com>]

On 2021/8/3 23:03, Eric Blake wrote:
> On Fri, Apr 02, 2021 at 11:52:25AM +0800, Zhenyu Ye wrote:
>> Hi all,
>>
>> commit 8dcd3c9b91 ("qemu-img: align result of is_allocated_sectors")
>> introduces block alignment when doing qemu-img convert. However, the
>> alignment is:
>>
>> 	s.alignment = MAX(pow2floor(s.min_sparse),
>>                       DIV_ROUND_UP(out_bs->bl.request_alignment,
>>                                    BDRV_SECTOR_SIZE));
>>
>> (where the default s.min_sparse is 8)
>> When the target device's bl.request_alignment is smaller than 4K, this
>> will cause additional write-zero overhead and makes the size of target
>> file larger.
>>
>> Is this as expected?  Should we change the MAX() to MIN()?
> 
> Yes it is expected, and no we shouldn't change it.  Even when a target
> advertises a bl.request_alignment of 512, our goal is to avoid needing
> read-modify-write cycles when that target is really on top of a 4k
> sector disk.  Writing extra 0s out to the 4k boundaries does not
> change the fact that allocation is in 4k chunks anyways, regardless of
> whether the disk supports smaller 512-byte reads.
> 

Thanks for your reply.

Zhenyu