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Re: Calculating swidth On A RAID6 Volume

To: Emmanuel Florac <eflorac@xxxxxxxxxxxxxx>
Subject: Re: Calculating swidth On A RAID6 Volume
From: Dave Chinner <david@xxxxxxxxxxxxx>
Date: Wed, 21 Jul 2010 16:54:29 +1000
Cc: xfs@xxxxxxxxxxx
In-reply-to: <20100720172704.609477f4@xxxxxxxxxxxxxxxxxxxx>
References: <3C5E029826A0704DB5998577FCFF46F0094DA1AD69@dagobah> <4C45AF09.5090901@xxxxxxxxxxxx> <20100720164444.4c1cef88@xxxxxxxxxxxxxxxxxxxx> <201007201700.51977@xxxxxx> <20100720172704.609477f4@xxxxxxxxxxxxxxxxxxxx>
User-agent: Mutt/1.5.20 (2009-06-14)
On Tue, Jul 20, 2010 at 05:27:04PM +0200, Emmanuel Florac wrote:
> Le Tue, 20 Jul 2010 17:00:51 +0200
> Michael Monnerie <michael.monnerie@xxxxxxxxxxxxxxxxxxx> écrivait:
> 
> > I'd say using su+sw is more future proof than swidth+sunit, as 4K
> > sector drives will become standard, and then using 512B units will be
> > outdated anyway, right?
> 
> Absolutely, and furthermore I'm wondering what's happening in the
> case where the drives have 4096 bytes blocks; I suppose then sunit
> should be expressed as a number of 4096 bytes blocks, and what about
> swidth ? the hell if I know :) This is also probably a nice little nest
> of coming filesystem bugs to be hatched :=)

No, the sky is not going to fall. ;)

The stripe unit is not related to sector size except for the fact
that sector size defines the minimum filesystem block size and the
stripe unit is an integer mutliple of the filesystem block size.
i.e.

mkfs.xfs -s size=512 -b size=512 -d sunit=1,swidth=1

is a valid configuration that gives single sector sunit/swidth
alignment. Hence sunit needs to be able to express sizes in
multiples of 512 bytes.

For a 4k sector size drive, the equivalent is:

mkfs.xfs -s size=4096 -b size=4096 -d sunit=8,swidth=8

and mkfs.xfs will ensure that any value of sunit that is not
a multiple of 8 (4k) will be rejected.....

Cheers,

Dave.
-- 
Dave Chinner
david@xxxxxxxxxxxxx

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