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List:       linux-xfs
Subject:    Re: Conversion Routine Help
From:       Dave Chinner <david () fromorbit ! com>
Date:       2011-12-29 21:54:12
Message-ID: 20111229215412.GM23662 () dastard
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On Thu, Dec 29, 2011 at 07:42:31PM +0530, Vijay Chauhan wrote:
> Hi list,
> 
> Can anyone please provide me links about the basic explanations of XFS
> basic blocks and filesystem logical block mapping and conversion
> routines?
> 
> I tried to understand from code but its not clear to me:
> #define XFS_FSB_TO_BB(mp,fsbno) ((fsbno) << (mp)->m_blkbb_log)

Convert FileSystem Blocks to Basic Blocks

FSBs are defined at mkfs time, BBs are always 512 bytes.

> #define XFS_BB_TO_FSB(mp,bb)    \
>        (((bb) + (XFS_FSB_TO_BB(mp,1) - 1)) >> (mp)->m_blkbb_log)

Convert BB to FSB, rounding up to the next FSB

> #define XFS_BB_TO_FSBT(mp,bb)   ((bb) >> (mp)->m_blkbb_log)

Convert BB to FSB, rounding down to the FSB containing the BB

> #define XFS_BB_FSB_OFFSET(mp,bb) ((bb) & ((mp)->m_bsize - 1))

Offset of the give BB within a FSB. e.g. if FSB = 4k = 8BB, then
"bb = 5" would return 5, "bb = 63" would return 7...

> lets consider the last one:
> #define XFS_BB_FSB_OFFSET(mp,bb) ((bb) & ((mp)->m_bsize - 1))
> 
> if basic block (512 byte size) number is 7 and m_bsize is 12

m_bsize is the FSB in BB, which will always be a power of 2.

> (considering FS block size 4096), then this will return 3 [e.g. ( 7 &

in that case, m_bsize = 8.

> 11) ]. then what does 3 means here? what offset value it is denoting?

It's the offset of the BB within the first partial FSB in the range
given. This was once used for sub-block zeroing needed by direct IO,
but is now stale code as the generic DIO layer does this zeroing.
Care to submit a patch to remove that macro?

Cheers,

Dave.
-- 
Dave Chinner
david@fromorbit.com

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