ČÁST I – Active Directory 18
00000000: EB 3C 90 4D 53 44 4F 53 – 35 2E 30 00 02 40 01 00 .<.MSDOS5.0..@..
00000010: 02 00 02 00 00 F8 FC 00 – 3F 00 40 00 3F 00 00 00 ........?.@.?...
00000020: 01 F0 3E 00 80 00 29 A8 – 8B 36 52 4E 4F 20 4E 41 ..>...)..6RNO NA
00000030: 4D 45 20 20 20 20 46 41 – 54 31 36 20 20 20 33 C0 ME FAT16 3.
00000040: 8E D0 BC 00 7C 68 C0 07 – 1F A0 10 00 F7 26 16 00 ....|h......&..
00000050: 03 06 0E 00 50 91 B8 20 – 00 F7 26 11 00 8B 1E 0B ....P.. ..&.....
00000060: 00 03 C3 48 F7 F3 03 C8 – 89 0E 08 02 68 00 10 07 ...H........h...
00000070: 33 DB 8F 06 13 02 89 1E – 15 02 0E E8 90 00 72 57 3.............rW
00000080: 33 DB 8B 0E 11 00 8B FB – 51 B9 0B 00 BE DC 01 F3 3.......Q.......
00000090: A6 59 74 05 83 C3 20 E2 – ED E3 37 26 8B 57 1A 52 .Yt... ...7&.W.R
000000A0: B8 01 00 68 00 20 07 33 – DB 0E E8 48 00 72 28 5B ...h. .3...H.r([
000000B0: 8D 36 0B 00 8D 3E 0B 02 – 1E 8F 45 02 C7 05 F5 00 .6...>....E.....
000000C0: 1E 8F 45 06 C7 45 04 0E – 01 8A 16 24 00 EA 03 00 ..E..E.....$....
000000D0: 00 20 BE 86 01 EB 03 BE – A2 01 E8 09 00 BE C1 01 . ..............
000000E0: E8 03 00 FB EB FE AC 0A – C0 74 09 B4 0E BB 07 00 .........t......
000000F0: CD 10 EB F2 C3 50 4A 4A – A0 0D 00 32 E4 F7 E2 03 .....PJJ...2....
00000100: 06 08 02 83 D2 00 A3 13 – 02 89 16 15 02 58 A2 07 .............X..
00000110: 02 A1 13 02 8B 16 15 02 – 03 06 1C 00 13 16 1E 00 ................
00000120: F7 36 18 00 FE C2 88 16 – 06 02 33 D2 F7 36 1A 00 .6........3..6..
00000130: 88 16 25 00 A3 04 02 A1 – 18 00 2A 06 06 02 40 3A ..%.......*...@:
00000140: 06 07 02 76 05 A0 07 02 – 32 E4 50 B4 02 8B 0E 04 ...v....2.P.....
00000150: 02 C0 E5 06 0A 2E 06 02 – 86 E9 8B 16 24 00 CD 13 ............$...
00000160: 0F 83 05 00 83 C4 02 F9 – CB 58 28 06 07 02 76 11 .........X(...v.
00000170: 01 06 13 02 83 16 15 02 – 00 F7 26 0B 00 03 D8 EB ..........&.....
00000180: 90 A2 07 02 F8 CB 42 4F – 4F 54 3A 20 43 6F 75 6C ......BOOT: Coul
00000190: 64 6E 27 74 20 66 69 6E – 64 20 4E 54 4C 44 52 0D dn’t find NTLDR.
000001A0: 0A 00 42 4F 4F 54 3A 20 – 49 2F 4F 20 65 72 72 6F ..BOOT: I/O erro
000001B0: 72 20 72 65 61 64 69 6E – 67 20 64 69 73 6B 0D 0A r reading disk..
000001C0: 00 50 6C 65 61 73 65 20 – 69 6E 73 65 72 74 20 61 .Please insert a
000001D0: 6E 6F 74 68 65 72 20 64 – 69 73 6B 00 4E 54 4C 44 nother disk.NTLD
000001E0: 52 20 20 20 20 20 20 00 – 00 00 00 00 00 00 00 00 R .........
000001F0: 00 00 00 00 00 00 00 00 – 00 00 00 00 00 00 55 AA ..............U.
Tabulky 1.7 a 1.8 ilustrují rozvržení BPB a rozšířeného BPB svazků FAT16. Vzorové
hodnoty korespondují s daty z předchozího příkladu.
Tabulka 1.7 Pole BPB svazků FAT16
00000010: 02 00 02 00 00 F8 FC 00 – 3F 00 40 00 3F 00 00 00 ........?.@.?...
00000020: 01 F0 3E 00 80 00 29 A8 – 8B 36 52 4E 4F 20 4E 41 ..>...)..6RNO NA
00000030: 4D 45 20 20 20 20 46 41 – 54 31 36 20 20 20 33 C0 ME FAT16 3.
00000040: 8E D0 BC 00 7C 68 C0 07 – 1F A0 10 00 F7 26 16 00 ....|h......&..
00000050: 03 06 0E 00 50 91 B8 20 – 00 F7 26 11 00 8B 1E 0B ....P.. ..&.....
00000060: 00 03 C3 48 F7 F3 03 C8 – 89 0E 08 02 68 00 10 07 ...H........h...
00000070: 33 DB 8F 06 13 02 89 1E – 15 02 0E E8 90 00 72 57 3.............rW
00000080: 33 DB 8B 0E 11 00 8B FB – 51 B9 0B 00 BE DC 01 F3 3.......Q.......
00000090: A6 59 74 05 83 C3 20 E2 – ED E3 37 26 8B 57 1A 52 .Yt... ...7&.W.R
000000A0: B8 01 00 68 00 20 07 33 – DB 0E E8 48 00 72 28 5B ...h. .3...H.r([
000000B0: 8D 36 0B 00 8D 3E 0B 02 – 1E 8F 45 02 C7 05 F5 00 .6...>....E.....
000000C0: 1E 8F 45 06 C7 45 04 0E – 01 8A 16 24 00 EA 03 00 ..E..E.....$....
000000D0: 00 20 BE 86 01 EB 03 BE – A2 01 E8 09 00 BE C1 01 . ..............
000000E0: E8 03 00 FB EB FE AC 0A – C0 74 09 B4 0E BB 07 00 .........t......
000000F0: CD 10 EB F2 C3 50 4A 4A – A0 0D 00 32 E4 F7 E2 03 .....PJJ...2....
00000100: 06 08 02 83 D2 00 A3 13 – 02 89 16 15 02 58 A2 07 .............X..
00000110: 02 A1 13 02 8B 16 15 02 – 03 06 1C 00 13 16 1E 00 ................
00000120: F7 36 18 00 FE C2 88 16 – 06 02 33 D2 F7 36 1A 00 .6........3..6..
00000130: 88 16 25 00 A3 04 02 A1 – 18 00 2A 06 06 02 40 3A ..%.......*...@:
00000140: 06 07 02 76 05 A0 07 02 – 32 E4 50 B4 02 8B 0E 04 ...v....2.P.....
00000150: 02 C0 E5 06 0A 2E 06 02 – 86 E9 8B 16 24 00 CD 13 ............$...
00000160: 0F 83 05 00 83 C4 02 F9 – CB 58 28 06 07 02 76 11 .........X(...v.
00000170: 01 06 13 02 83 16 15 02 – 00 F7 26 0B 00 03 D8 EB ..........&.....
00000180: 90 A2 07 02 F8 CB 42 4F – 4F 54 3A 20 43 6F 75 6C ......BOOT: Coul
00000190: 64 6E 27 74 20 66 69 6E – 64 20 4E 54 4C 44 52 0D dn’t find NTLDR.
000001A0: 0A 00 42 4F 4F 54 3A 20 – 49 2F 4F 20 65 72 72 6F ..BOOT: I/O erro
000001B0: 72 20 72 65 61 64 69 6E – 67 20 64 69 73 6B 0D 0A r reading disk..
000001C0: 00 50 6C 65 61 73 65 20 – 69 6E 73 65 72 74 20 61 .Please insert a
000001D0: 6E 6F 74 68 65 72 20 64 – 69 73 6B 00 4E 54 4C 44 nother disk.NTLD
000001E0: 52 20 20 20 20 20 20 00 – 00 00 00 00 00 00 00 00 R .........
000001F0: 00 00 00 00 00 00 00 00 – 00 00 00 00 00 00 55 AA ..............U.
Tabulky 1.7 a 1.8 ilustrují rozvržení BPB a rozšířeného BPB svazků FAT16. Vzorové
hodnoty korespondují s daty z předchozího příkladu.
Tabulka 1.7 Pole BPB svazků FAT16
posted by rufus at 12:08 AM
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