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bytes per inode question

Dear experts,

My question is about bytes-per-inode when formatting a partition. The followings are what I did:

#mke2fs -i 8192 /dev/hda3
#mount /dev/hda3 /tmp/mydir
#cd /tmp/mydir
#mkdir hello
#ls -al
drwxr-xr-x 2 root root 4096 Oct 23 16:28 hello
#

I already indicated one inode points to 8192 (2 blocks) bytes. Since a directory uses one inode, I suppose the size of the directory should be 8192. But it seemes that I am wrong. Why the directory size is 4096?

Thanks in advance

ltsun

ASKER CERTIFIED SOLUTION

MFCRich

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kyrmit

Hi,
there are currently two thoughts I have here:

the inode size defines the maximum space that can be used within an inode. If the directory listing is smaller than the required space, it may not be needing all of the 8192 bytes initially. Once the directory index exceeds the required size, the structure gets expanded.

and secondly, according to my quick browse through the sources, I doubt somehow that you can allocate inodes of this size . The maximum (according to the sources in 2.2.19 kernel ) is 4096 bytes. Since mke2fs is not a kernel tool, it may have attempted to do more.

Both of which may account independantly. You may want to check the block size tune2fs -l /dev/hda3 too see if you actually got 8192 or 4096.

K.

MFCRich

You can not change the size of an inode without modifying kernel source code. Inode size is fixed! The 'i' switch to the mke2fs command affects the _number_ of inodes, not their size.

ltsun

ASKER

MFCRich,

I got the idea. Thanks very much.

ltsun