On 16-Oct-09, at 08:45, Vyacheslav Dubeyko wrote:
Let's try to make raw partition with 8 Gb size by fdisk and then to
format it (for example, mkfs.ext4 -b 4096 -L ext4_4K_8G /dev/sdb1).
If we have 2098482 block count on volume with 4 Kb block size and
flex block group size as 16 then we will have 65 groups on volume.
The last group (that has 1329 blocks) will be the first and sole
group in last flex group. Current mke2fs code makes such allocation
scheme in last group: Block bitmap at 2097152 (+0), Inode bitmap at
2097168 (+16), Inode table at 8626 - 9130. The inode table of the
last group will be placed at the volume begin because of we can't
allocate sufficient block count for all inode tables in flex group.
I offer the patch for mke2fs utility:
diff --git a/lib/ext2fs/alloc_tables.c b/lib/ext2fs/alloc_tables.c
index 8547ad6..4639527 100644
--- a/lib/ext2fs/alloc_tables.c
+++ b/lib/ext2fs/alloc_tables.c
@@ -181,10 +181,15 @@ errcode_t
ext2fs_allocate_group_table(ext2_filsys fs, dgrp_t group,
blk_t prev_block = 0;
if (group && fs->group_desc[group-1].bg_inode_table)
prev_block = fs->group_desc[group-1].bg_inode_table;
+ int requsted_size = 0;
This C++ style declaration needs to go at the beginning of the function.
+ if ((group+1) == fs->group_desc_count &&
+ (fs->group_desc_count % flexbg_size) == 1)
+ requsted_size = fs->inode_blocks_per_group;
+ else
+ requsted_size = fs->inode_blocks_per_group * rem_grps;
group_blk = flexbg_offset(fs, group, prev_block, bmap,
flexbg_size * 2,
- fs->inode_blocks_per_group *
- rem_grps,
+ requsted_size,
fs->inode_blocks_per_group);
last_blk = ext2fs_group_last_block(fs, last_grp);
}
I think the rest of this looks reasonable.
Cheers, Andreas
--
Andreas Dilger
Sr. Staff Engineer, Lustre Group
Sun Microsystems of Canada, Inc.
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