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Directory Dump-System Programming-Lecture Notes, Study notes of System Programming

Bharat Ratna Dr. B. R. Ambedkar UniversitySystem Programming

This lecture handout is for System Programming course. It was provided by Prof. Indubhushan Vijayabhas at Ambedkar University, Delhi. It includes: Directory, Dump, Slide, Block, Read, Root, Access, Larger, File, Content, Reset, Table

Typology: Study notes

2011/2012

Uploaded on 08/07/2012

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Lecture # 34
Here is another example which examines the contents of a file for a FAT 16 system.
Firstly the DPB is read as shown In the following slide.
Another Example with FAT16 bit System
DPB Dump
-a
13AD:0100 mov ah,32
13AD:0102 int 21
13AD:0104
-p
AX=3200 BX=0000 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000
DS=13AD ES=13AD SS=13AD CS=13AD IP=0102 NV UP EI PL NZ NA PO NC
13AD:0102 CD21 INT 21
-p
AX=3200 BX=13D2 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 DI=0000
DS=00A7 ES=13AD SS=13AD CS=13AD IP=0104 NV UP EI PL NZ NA PO NC
13AD:0104 D3E3 SHL BX,CL
-d a7:13d2
00A7:13D0 05 05 00 02 07 03-08 00 02 00 02 C0 01 B4 ..............
00A7:13E0 CB CC 00 A0 01 56 34 12-00 F8 0A FF FF FF FF 00 .....V4.........
00A7:13F0 00 98 CB 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
00A7:1400 0 0 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
00A7:1410 0 0 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
00A7:1420 0 0 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
00A7:1430 0 0 00 00 00 00 00 00 0C-00 00 80 00 B7 13 10 00 ................
00A7:1440 DF 12 4D 5A 9A 00 29 00-00 00 20 00 C5 00 FF FF ..MZ..)... .....
00A7:1450 A7 05 ..
-q
Once the DPB has been read the blocks reserved for root directory are determined and are
then read to get the contents of the root directory.
Directory Dump
-l 1000 5 1a 0 20
-d 1000 4000
13AD:1000 4E 45 57 20 56 4F 4C 55-4D 45 20 08 00 00 00 00 N EW VOLUME .....
13AD:1010 00 00 00 00 00 00 61 76-2D 33 00 00 00 00 00 00 . .....av-3......
13AD:1020 41 52 00 65 00 63 00 79-00 63 00 0F 00 21 6C 00 AR.e.c. y.c...!l.
13AD:1030 65 00 64 00 00 00 FF FF-FF FF 00 00 FF FF FF FF e .d.............
13AD:1040 52 45 43 59 43 4C 45 44-20 20 20 16 00 4E 79 5E RECYCLE D ..Ny^
13AD:1050 2F 33 2F 33 00 00 7A 5E-2F 33 02 00 00 00 00 00 / 3/3..z^/3......
13AD:1060 42 20 00 49 00 6E 00 66-00 6F 00 0F 00 72 72 00 B .I.n. f.o...rr.
13AD:1070 6D 00 61 00 74 00 69 00-6F 00 00 00 6E 00 00 00 m .a.t.i.o...n...
13AD:1080 01 53 00 79 00 73 00 74-00 65 00 0F 00 72 6D 00 .S.y.s. t.e...rm.
13AD:1090 20 00 56 00 6F 00 6C 00-75 00 00 00 6D 00 65 00 .V.o.l .u...m.e.
13AD:10A0 53 59 53 54 45 4D 7E 31-20 20 20 16 00 4E 79 5E SYSTEM~ 1 ..Ny^
13AD:10B0 2F 33 2F 33 00 00 7A 5E-2F 33 03 00 00 00 00 00 / 3/3..z^/3......
13AD:10C0 41 64 00 50 00 62 00 31-00 2E 00 0F 00 6A 74 00 Ad.P.b. 1.....jt.
13AD:10D0 78 00 74 00 00 00 FF FF-FF FF 00 00 FF FF FF FF x .t.............
13AD:10E0 44 50 42 31 20 20 20 20-54 58 54 20 00 57 81 69 DPB1 TXT .W.i
13AD:10F0 36 33 36 33 00 00 07 78-36 33 0A 00 8A 06 00 00 6363...x63......
13AD:1100 44 50 42 32 20 20 20 20-54 58 54 20 18 12 AE 69 DPB2 TXT ...i
13AD:1110 36 33 36 33 00 00 03 75-36 33 0B 00 5F 06 00 00 6363...u63.._...
13AD:1120 46 49 52 53 54 20 20 20-20 20 20 10 08 6F ED 56 FIRST ..o.V
13AD:1130 3C 33 3C 33 00 00 EE 56-3C 33 0C 00 00 00 00 00 <3<3...V<3......
13AD:1140 53 45 43 4F 4E 44 20 20-20 20 20 10 08 50 EF 56 SECOND ..P.V
13AD:1150 3C 33 3C 33 00 00 F0 56-3C 33 12 00 00 00 00 00 <3<3...V<3......
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Lecture # 34

Here is another example which examines the contents of a file for a FAT 16 system.

Firstly the DPB is read as shown In the following slide.

Another Example with FAT16 bit System

DPB Dump

-a 13AD:0100 mov ah, 13AD:0102 int 21 13AD:

-p AX=3200 BX=0000 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 D I= DS=13AD ES=13AD SS=13AD CS=13AD IP=0102 NV UP EI PL NZ NA PO NC 13AD:0102 CD21 INT 21

-p AX=3200 BX=13D2 CX=0000 DX=0000 SP=FFEE BP=0000 SI=0000 D I= DS=00A7 ES=13AD SS=13AD CS=13AD IP=0104 NV UP EI PL NZ NA PO NC 13AD:0104 D3E3 SHL BX,CL

-d a7:13d 00A7:13D0 05 05 00 02 07 03-08 00 02 00 02 C0 01 B .............. 00A7:13E0 CB CC 00 A0 01 56 34 12-00 F8 0A FF FF FF FF 00 .....V4......... 00A7:13F0 00 98 CB 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 00A7:1400 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 00A7:1410 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 00A7:1420 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 00A7:1430 00 00 00 00 00 00 00 0C-00 00 80 00 B7 13 10 00 ................ 00A7:1440 DF 12 4D 5A 9A 00 29 00-00 00 20 00 C5 00 FF FF ..MZ..)... ..... 00A7:1450 A7 05 .. -q

Once the DPB has been read the blocks reserved for root directory are determined and are

then read to get the contents of the root directory.

Directory Dump

-l 1000 5 1a0 20

-d 1000 4000 13AD:100 0 4E 45 57 20 56 4F 4C 55-4D 45 20 08 00 00 00 00 N EW VOL UME .. ... 13AD:101 0 00 00 00 00 00 00 61 76-2D 33 00 00 00 00 00 00. .....a v-3... ... 13AD:102 0 41 52 00 65 00 63 00 79-00 63 00 0F 00 21 6C 00 A R.e.c. y.c... !l. 13AD:103 0 65 00 64 00 00 00 FF FF-FF FF 00 00 FF FF FF FF e .d.... ...... ... 13AD:104 0 52 45 43 59 43 4C 45 44-20 20 20 16 00 4E 79 5E R ECYCLE D .. Ny^ 13AD:105 0 2F 33 2F 33 00 00 7A 5E-2F 33 02 00 00 00 00 00 / 3/3..z ^/3... ... 13AD:106 0 42 20 00 49 00 6E 00 66-00 6F 00 0F 00 72 72 00 B .I.n. f.o... rr. 13AD:107 0 6D 00 61 00 74 00 69 00-6F 00 00 00 6E 00 00 00 m .a.t.i .o...n ... 13AD:108 0 01 53 00 79 00 73 00 74-00 65 00 0F 00 72 6D 00. S.y.s. t.e... rm. 13AD:109 0 20 00 56 00 6F 00 6C 00-75 00 00 00 6D 00 65 00 .V.o.l .u...m .e. 13AD:10A 0 53 59 53 54 45 4D 7E 31-20 20 20 16 00 4E 79 5E S YSTEM~ 1 .. Ny^ 13AD:10B 0 2F 33 2F 33 00 00 7A 5E-2F 33 03 00 00 00 00 00 / 3/3..z ^/3... ... 13AD:10C 0 41 64 00 50 00 62 00 31-00 2E 00 0F 00 6A 74 00 A d.P.b. 1..... jt. 13AD:10D 0 78 00 74 00 00 00 FF FF-FF FF 00 00 FF FF FF FF x .t.... ...... ... 13AD:10E0 44 50 42 31 20 20 20 20-54 58 54 20 00 57 81 69 DPB1 TXT. W.i 13AD:10F0 36 33 36 33 00 00 07 78-36 33 0A 00 8A 06 00 00 6363...x63...... 13AD:1100 44 50 42 32 20 20 20 20-54 58 54 20 18 12 AE 69 DPB2 TXT ...i 13AD:1110 36 33 36 33 00 00 03 75-36 33 0B 00 5F 06 00 00 6363...u63.._... 13AD:1120 46 49 52 53 54 20 20 20-20 20 20 10 08 6F ED 56 FIRST .. o.V 13AD:1130 3C 33 3C 33 00 00 EE 56-3C 33 0C 00 00 00 00 00 <3<3...V<3...... 13AD:1140 53 45 43 4F 4E 44 20 20-20 20 20 10 08 50 EF 56 SECOND ..P.V 13AD:1150 3C 33 3C 33 00 00 F0 56-3C 33 12 00 00 00 00 00 <3<3...V<3......

The root directory contains a collection of FCBs. The FCB for the file in question is

searched from where the first cluster of the file can be get.

Cont…

13AD:12C0 43 50 41 50 45 52 20 20-54 58 54 20 00 9D 7D 6F CPAPER TXT ..}o 13AD:12D0 3C 33 3C 33 00 00 60 05-72 28 29 00 4F 8F 00 00 <3<3..`.r().O... 13AD:12E0 E5 6D 00 65 00 6E 00 74-00 2E 00 0F 00 9F 74 00. m.e.n.t......t. 13AD:12F0 78 00 74 00 00 00 FF FF-FF FF 00 00 FF FF FF FF x.t............. 13AD:1300 E5 4E 00 65 00 77 00 20-00 54 00 0F 00 9F 65 00. N.e.w. .T....e. 13AD:1310 78 00 74 00 20 00 44 00-6F 00 00 00 63 00 75 00 x.t. .D.o...c.u. 13AD:1320 E5 45 57 54 45 58 7E 31-54 58 54 20 00 32 09 73 .EWTEX~1TXT .2.s 13AD:1330 3C 33 3C 33 00 00 0A 73-3C 33 00 00 00 00 00 00 <3<3...s<3...... 13AD:1340 54 45 53 54 20 20 20 20-54 58 54 20 18 32 09 73 TEST TXT .2.s 13AD:1350 3C 33 3C 33 00 00 17 73-3C 33 45 00 27 00 00 00 <3<3...s<3E.'...

File Contents File Size = 39D =27H File Cluster # = 45H File Sec # = (45H – 2 )*8 + 01C0 = 3D8H

After calculating the sector number for the cluster the contents of the file can be accessed

by reading all the blocks within the cluster. In this way only the starting cluster will be

read. If the file contains a number of cluster the subsequent clusters numbers within the

file chain can be accessed from the FAT.

File Dump

-l 1000 5 3d8 8

-d 1000

13AD:1000 74 68 69 73 20 69 73 20-61 20 74 65 73 74 20 74 this is a test t 13AD:1010 65 78 74 20 66 69 6C 65-20 66 6F 72 20 31 36 20 ext file for 16 13AD:1020 62 69 74 20 46 41 54 00-00 00 00 00 00 00 00 00 bit FAT......... 13AD:1030 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 13AD:1040 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 13AD:1050 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 13AD:1060 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 13AD:1070 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ -q

File Organization

R

R

12 – bit

Start Cluster

FCB

Cont…

EOF

2^

Above slides show how a cluster chain for a file is maintained in the FAT. The first

cluster number is in the FCB. Subsequent clusters will be accessed from the FAT using

the previous cluster number as index to look up into the FAT for the next cluster number.

A FAT theoretically will contain 2

n

entries where n is 12 for FAT 12 and 16 for FAT16.

But all the entries are not used some of the entries are reserved following slide shows its

detail.

Unused FAT Entries

  • Reserved Entries = FF0H ~ FF6H
  • EOF value = FF7H ~ FFFH
  • First Two Clusters = 0,
  • Free Cluster = 0
  • Max. range of Cluster # = 2 ~ FEFH
  • Total # of Clusters of FAT12 = FEEH

Cluster Size Determination

tempof = size of disk / no. of entries in FAT (FEEH)

if ( temp > 32768)

use higher FAT16 or FAT

else

{

choose the nearest value + temp greater than

temp, which is a power of 2,

Set this to be the Cluster size = bytes

Size of Cluster in Blocks =

Cluster size in Bytes / Bytes per Block

}

No. of Entries of FAT =

No. of Blocks in disk in User Data Area / Size

of Cluster in Blocks.

There can various volume with various sizes with FAT12 or FAT16. The number of

entries for FAT 12 or FAT16 are limited then the question arises how can a certain

volume with moderate space and another volume with large space can be managed by the

same FAT system. The answer is that the number of entries might be same but the size of

cluster may be different. The cluster size can vary from 512 bytes to 32K in powers of 2