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Date:	Thu, 22 Feb 2007 16:25:48 +0000
From:	Jörn Engel <joern@...ybastard.org>
To:	Juan Piernas Canovas <piernas@...ec.um.es>
Cc:	Sorin Faibish <sfaibish@....com>,
	kernel list <linux-kernel@...r.kernel.org>
Subject: Re: [ANNOUNCE] DualFS: File System with Meta-data and Data Separation

On Thu, 22 February 2007 05:30:03 +0100, Juan Piernas Canovas wrote:
> 
> DualFS writes meta-blocks in variable-sized chunks that we call partial 
> segments. The meta-data device, however, is divided into segments, which 
> have the same size. A partial segment can be as large a a segment, but a 
> segment usually has more that one partial segment. Besides, a partial 
> segment can not cross a segment boundary.

Sure, that's a fairly common approach.

> A partial segment is a transaction unit, and contains "all" the blocks 
> modified by a file system operation, including indirect blocks and i-nodes 
> (actually, it contains the blocks modified by several file system 
> operations, but let us assume that every partial segment only contains the 
> blocks modified by a single file system operation).
> 
> So, the above figure is as follows in DualFS:
> 
>  Before:
>  Segment 1: [some data] [ D0 D1 D2 I ] [more data]
>  Segment 2: [             some data              ]
>  Segment 3: [               empty                ]
> 
> If the datablock D0 is modified, what you get is:
> 
>  Segment 1: [some data] [  garbage   ] [more data]
>  Segment 2: [             some data              ]
>  Segment 3: [ D0 D1 D2 I ] [       empty         ]

You have fairly strict assumptions about the "Before:" picture.  But
what happens if those assumptions fail.  To give you an example, imagine
the following small script:

$ for i in `seq 1000000`; do touch $i; done

This will create a million dentries in one directory.  It will also
create a million inodes, but let us ignore those for a moment.  It is
fairly unlikely that you can fit a million dentries into [D0], so you
will need more than one block.  Let's call them [DA], [DB], [DC], etc.
So you have to write out the first block [DA].

 Before:
Segment 1: [some data] [ DA D1 D2 I ] [more data]
Segment 2: [             some data              ]
Segment 3: [               empty                ]

If the datablock D0 is modified, what you get is:

Segment 1: [some data] [  garbage   ] [more data]
Segment 2: [             some data              ]
Segment 3: [ DA D1 D2 I ] [       empty         ]

That is exactly your picture.  Fine.  Next you write [DB].

Before: see above
After:
Segment 1: [some data] [  garbage   ] [more data]
Segment 2: [             some data              ]
Segment 3: [ DA][garbage] [ DB D1 D2 I ] [ empty]

You write [DC].  Note that Segment 3 does not have enough space for
another partial segment:

Segment 1: [some data] [  garbage   ] [more data]
Segment 2: [             some data              ]
Segment 3: [ DA][garbage] [ DB][garbage] [wasted]
Segment 4: [ DC D1 D2 I ] [       empty         ]

You write [DD] and [DE]:
Segment 1: [some data] [  garbage   ] [more data]
Segment 2: [             some data              ]
Segment 3: [ DA][garbage] [ DB][garbage] [wasted]
Segment 4: [ DC][garbage] [ DD][garbage] [wasted]
Segment 5: [ DE D1 D2 I ] [       empty         ]

And some time later you even have to switch to a new indirect block, so
you get before:

Segment n  : [ DX D1 D2 I ] [       empty         ]

After:

Segment n  : [ DX D1][garb] [ DY DI D2 I ] [ empty]

What you end up with after all this is quite unlike you "Before"
picture.  Instead of this:

>  Segment 1: [some data] [ D0 D1 D2 I ] [more data]

You may have something closer to this:

> >Segment 1: [some data] [   D1  ] [more data]
> >Segment 2: [some data] [   D0  ] [more data]
> >Segment 3: [some data] [   D2  ] [more data]

You should try the testcase and look at a dump of your filesystem
afterwards.  I usually just read the raw device in a hex editor.

Jörn

-- 
Beware of bugs in the above code; I have only proved it correct, but
not tried it.
-- Donald Knuth
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