Re: [Question] qemu-img convert block alignment

[Eric Blake <eblake@redhat.com>]

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.

-- 
Eric Blake, Principal Software Engineer
Red Hat, Inc.           +1-919-301-3266
Virtualization:  qemu.org | libvirt.org