Hi, 2010/12/27 Vladimir 'φ-coder/phcoder' Serbinenko <phco...@gmail.com>: > On 12/26/2010 10:15 PM, ehem+g...@m5p.com wrote: >> Quite simple, the disk slice scheme detection routines vary in the >> quality of their detection. In particular, the MSDOS-style detection is >> *extremely* brittle. The only even mildly distinguishing characteristic >> it finds is ensuring only the msb of the boot-flag byte is set. The other >> thing it looks for is the 0xAA55 signature, but that is merely a signal >> to PC-BIOSes that the disk is bootable; as such *any* bootable disk for a >> IBM-PC will have that signature, whether or not it is actually using the >> MSDOS-style header. A 1 in 65536 chance of a false positive is bad.
If a disk is bootable then it should have the 0xAA55 signature, regardless of whether or not there's a partition table. If a disk is *not* bootable then it should *not* have the 0xAA55 signature, regardless of whether or not there's a partition table. Lots of OSs (including Linux/fdisk) have the same stupid bug where they put the 0xAA55 signature on pure data disks where it doesn't belong. The worst case here is that the computer crashes during boot if the disk ever becomes the "first" disk (rather than the BIOS moving on to the next type of bootable device, and/or rather than the user getting a "No OS found" BIOS error message), so it's usually just an annoying bug (and not a serious bug, unless someone is perverted enough to attempt to use a bootable CD-ROM or something as a type of fail-safe in case their first drive goes missing), but it's still "wrong". To check if there is/isn't an "MBR partition table", first check for GPT. If there's no GPT then; for each (potential) MBR partition table entry that isn't "type = unused" check for "(boot-flag & 0x7F == 0)" then check if the starting CHS address is sane (e.g. not an invalid sector, and with work-arounds for "unrepresentable") and if the ending CHS address is sane (e.g. not invalid and not less than or equal to the starting CHS, and with work-arounds for "unrepresentable") and then do the same for the starting and ending LBA addresses. Then you could check for overlapping partitions. If a partition is marked as "active" then you could check for the 0xAA55 signature in the first sector of that partition, but that is the only case where checking the 0xAA55 signature makes any sense. > Actually 1 in 2^(7*4+16) =2^44 in you take into account the both checks > and consider every possible sector equiprobable. Actually, if you take into account that the second check is worthless and/or wrong; then for the current code; if there's 4 (assumed) "usable" partitions there's 1 chance in 4*127 of thinking there's a partition table when there isn't, and if there's only one (assumed) "usable" partition then it drops to 1 chance in 127; and if the MBR contains random data the former is much more likely than the latter (I was too lazy to do the calculation properly). Cheers, Brendan _______________________________________________ Grub-devel mailing list Grub-devel@gnu.org http://lists.gnu.org/mailman/listinfo/grub-devel
