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| Subject: | [Full-disclosure] Re: [VulnWatch] Microsoft Windows NTFS Information Disclosure |
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| Date: | Thu, 30 Jun 2005 14:17:07 -0700 (PDT) |
mattmurphy@kc.rr.com (Matthew Murphy) at Jun 30, 2005 12:01:59 PM wrote:
However, an apparent error in the NTFS driver's code causes the file system to incorrectly assign disk blocks to files before they have been initialized. Following a recovery from a system shutdown, uninitialized data may be visible in files from previously allocated disk blocks.
As far as I know, _every_ major Operating System has the same vulnerability. I do _NOT_ know of any Operating System that "zero's" each newly-allocated block/sector/track/cylinder of disk-space when allocating a "new" file, whether on disk, or on magnetic tape, or on removable media. IBM AIX? No. IBM z/VM? No. IBM z/OS? No. IBM OS/2? No. HP/UX? No. Linux? No. MS DOS? No. MS Windows? No. The "cost" (in elapsed time, due to I/O waits) would be an unacceptable performance-penalty. The "benefit" would be that one would never need to 'FORMAT' a 3.5" disk; the O.S. would do it for you, the first time that you tried to write to it. I would hate to work with an O.S. that would automatically "write" on a magnetic tape to completely fill the tape, _before_ allowing "backup" software to rewrite useful information onto a tape. Backups would take _double_ the elapsed-time to complete ?!?! Unacceptable. P.S. If you're looking for a "vulnerability", consider how a logical record is updated within a file-system. Typically, an "update-in-place" mechanism is not used. Instead, when a logical record is to be updated, the physical block containing that logical record is read from the disk, the logical record is updated in a buffer, and the file-system finds an "unused" physical block on the disk, and copies from the buffer to that new block. If there is a "pointer-block" that points to all the "data" blocks in the file, then that pointer-block is read from the disk, updated in the buffer, and, like the updated "data" logical record, written to an "unused" block on the disk. If there is a "file-directory" structure on the disk, an update occurs to that structure, and the updated part of that file-directory is written to an "unused" block on disk. So, updating just _one_ logical record can cause many I/O operations, and the consumption of many of those "previously-unused" blocks on the disk. However, note that the "dead" block(s) are not "zero'd" during the update. Why? Because, if the operating-system crashes during these updates, the original "file-system" structure is still present on the disk, and the original "pointer-blocks" and "data-blocks" are still present, i.e., after the crash, the "original" file has returned, and the file-system in a "consistent" status, and any of the "incomplete" updates have been "rolled-back". _______________________________________________ Full-Disclosure - We believe in it. Charter: http://lists.grok.org.uk/full-disclosure-charter.html Hosted and sponsored by Secunia - http://secunia.com/
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