Re: How to find the inodes in XFS

[Felipe Monteiro de Carvalho <felipemonteiro.carvalho@xxxxxxxxx>]

Oh, well, ...now my code works great for low inode numbers, but it
doesn't work at all for high inode numbers. =)

If inode is the root one 0x80 or similar small numbers which as 0x83
it will map correctly to the block 8, which means position 0x8000 +
internal offset inside the block

But in high inodes, for example 0x204B87 and 0x204B80 it will split
the inode number like this:
AG=2
Block inside AG Nr=0x4B8

And since XFS_INO_TO_FSB just recombines back together the parts:

#define XFS_INO_TO_FSB(mp,i) \
XFS_AGB_TO_FSB(mp, XFS_INO_TO_AGNO(mp,i), XFS_INO_TO_AGBNO(mp,i))

#define XFS_AGB_TO_FSB(mp,agno,agbno) \
(((xfs_fsblock_t)(agno) << (mp)->m_sb.sb_agblklog) | (agbno))

I get as the result from XFS_INO_TO_FSB 0x204B8, which would mean
position 0x204B8000, but this is a wrong position =(

I opened the partition with a hex viewer and the correct position is
block 0x1521A which means position 0x1521A000 + offset inside block

My superblock has these values:

XFS_INO_AGINO_BITS = mp^.m_agino_log = 20
XFS_INO_OFFSET_BITS= m_sb^.sb_inopblog = 4
XFS_INO_AGBNO_BITS = m_sb^.sb_agblklog = 16
XFS_INO_MASK = FFFF
sb_agcount = 4

Any ideas?

Maybe my flaw is that to get the disk position I simply multiply the
block number by the size of a block ... I tryed to use
XFS_FSB_TO_DADDR instead but it gives 64 for block nr 8, which doesn't
make much sense, I'd expect 0x8000

thanks!
-- 
Felipe Monteiro de Carvalho