DE3
FILE INFORMATION
FILENAME(S): DE3
FILE TYPE(S): PRG
FILE SIZE: 11.1K
FIRST SEEN: 2025-11-30 17:31:22
APPEARS ON: 1 disk(s)
FILE HASH
ea57577124ffea0e15e092057ecece5c04007707cbab87d3ccbb3bb6a529e05f
FOUND ON DISKS (1 DISKS)
| DISK TITLE | FILENAME | FILE TYPE | COLLECTION | TRACK | SECTOR | ACTIONS |
|---|---|---|---|---|---|---|
| CDU FEBRUARY 90 | DE3 | PRG | UK CD | 24 | 6 | DOWNLOAD FILE |
FILE CONTENT & ANALYSIS
> DETECTED: COMMODORE BASIC PROGRAM (LOAD ADDRESS $0801)
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00002B00: 11 9D 9D 9D BC 12 A2 92 20 22 2C 22 12 A2 A2 92 |........ ","....|
00002B10: A1 11 9D 9D 9D 20 12 BF 92 20 11 9D 9D 9D 20 A1 |..... ... .... .|
00002B20: 20 11 9D 9D 9D 20 BE 20 22 00 6C 33 E9 00 83 22 | .... . ".l3..."|
00002B30: 12 BF A2 92 BB 11 9D 9D 9D BF A2 BE 11 9D 9D 9D |................|
00002B40: A1 20 A1 11 9D 9D 9D BC 12 A2 92 20 22 2C 22 12 |. ......... ",".|
00002B50: BF A2 92 BB 11 9D 9D 9D BF A2 A1 11 9D 9D 9D BB |................|
00002B60: 20 A1 11 9D 9D 9D BC 12 A2 92 20 22 00 AF 33 EA | ......... "..3.|
00002B70: 00 83 22 AC BF 20 11 9D 9D 9D 12 BC 92 A2 A1 11 |..".. ..........|
00002B80: 9D 9D 9D A1 20 A1 11 9D 9D 9D BE 20 BE 22 2C 22 |.... ...... .","|
00002B90: 12 AC A2 92 BB 11 9D 9D 9D 12 BC 92 A2 BE 11 9D |................|
00002BA0: 9D 9D A1 20 A1 11 9D 9D 9D 12 A2 A2 92 20 22 00 |... ......... ".|
00002BB0: F2 33 EB 00 83 22 12 BF A2 92 BB 11 9D 9D 9D A1 |.3..."..........|
00002BC0: 20 20 11 9D 9D 9D A1 20 BB 11 9D 9D 9D BC 12 A2 | ..... ........|
00002BD0: 92 20 22 2C 22 12 AC 92 BF 20 11 9D 9D 9D A1 20 |. ",".... ..... |
00002BE0: A1 11 9D 9D 9D A1 AC BE 11 9D 9D 9D 12 A2 92 BE |................|
00002BF0: 20 22 00 37 34 EC 00 83 22 12 AC A2 92 BE 11 9D | ".74...".......|
00002C00: 9D 9D 12 BC 92 BB 20 11 9D 9D 9D A1 20 20 11 9D |...... ..... ..|
00002C10: 9D 9D 12 A2 A2 92 BE 22 2C 22 12 AC A2 92 BE 11 |.......","......|
00002C20: 9D 9D 9D 12 BC 92 BB 20 11 9D 9D 9D A1 20 20 11 |....... ..... .|
00002C30: 9D 9D 9D BE 20 20 22 00 00 00 |.... "... |
......A.A.1:.A.1.."DE.CODE",8,1.[...KY.4
9235:L.49152:.H$(15):C(0).1:C(1).15:C(2)
.12:C(3).11......53280,5:.53281,11:."...
.. ONCE A DISK HAS BEEN FORMATTED BY TH
E".R....". COMMODORE 64 IT CONTAINS 3
5 TRACKS,"......". EACH ONE HAVING BE
TWEEN 17 AND 21".2....". SECTORS (OR B
LOCKS). ALTOGETHER THERE".B....". ARE 6
83 BLOCKS PER DISK, 664 OF WHICH"......
". CAN BE USED FREELY BY YOU FOR STORIN
G".B....". WHATEVER SORT OF DATA YOU WI
SH. EACH"......". TIME A PROGRAM IS
SAVED, THE DRIVE"."....". NEEDS TO KNOW
AS QUICKLY AS POSSIBLE".R....". WHICH
SECTORS ARE FREE, WITHOUT HAVING"......
". TO SEARCH THROUGH THE ENTIRE DISK.":.
I.0.15:.H$(I).V...H$(I).H$(I)."...":.:.K
Y:.".. IT IS FOR THIS REASON THAT SECTOR
ZERO"......". OF THE DIRECTORY TRACK I
S HOME TO THE".6....". MOST IMPORTANT
BLOCK OF 140 BYTES".F....". ANYWHERE
ON THE DISK. IT IS THE BLOCK"......". A
LLOCATION MAP AND IT KEEPS A RECORD".F
....". OF EXACTLY WHICH OF THE 664 SE
CTORS"......". ARE AVAILABLE FOR USE.
EACH TRACK ON".&....". THE DISK IS REPRE
SENTED BY FOUR BYTES,".V....". THE FIRS
T BEING THE TOTAL NUMBER OF"......". F
REE SECTORS ON THAT TRACK, THE OTHERS"..
....". STORING THE INFORMATION FOR THE B
REAK-"......". DOWN OF EXACTLY WHICH SEC
TORS ARE FREE"......KY:.".. TO DEMONSTR
ATE THE BLOCK ALLOCATION".K....". MAP
- BAM FOR SHORT - YOU SHOULD NOW".{...
.". INSERT A DISK OF YOUR CHOICE. THE DA
TA"......". FROM THE BAM ON THAT DISK
WILL THEN"......". BE USED IN THE RE
ST OF THIS DEMO."......". IT IS IMPORT
ANT THAT YOU DO NOT REMOVE".N. ..". THAT
DISK UNTIL THE DEMONSTRATION HAS":.".
FINISHED.":.KY...!.."..TRACK 18, SECTOR
0.":."..";:.15,8,15:.8,8,8,"#"..."..15,"
U1 8 0 18 0":.I.0.7:.K.0.1:.J.1.16...#..
#8,A$:A.F(A$.G(0)):H$."":.M.1.2:D..(A.16
)...$.H$.J("0123456789ABCDEF",1.A.D.16,1
).H$:A.D:..:.%..199,.(.(I.2..(I.2))..(J.
2..(J.2))).R.&..H$;:.:.:." ";:.K,I.}.'..
"..THE ABOVE DATA REPRESENTS THE 256"...
(.." BYTES OF INFORMATION STORED AT THE"
.T.).." START OF THE DIRECTORY TRACK. LE
T'S"...*.." WORK THROUGH IT, HIGHLIGHTIN
G THE".5.+.." IMPORTANT SECTIONS.":.KY:.
2023,32:.L,720,960,11.U.,..L,80,720,0:."
...... THE FIRST TWO BYTES OF ANY SECTOR
ARE"...-.." POINTERS AND TELL THE DRIVE
WHICH"......" TRACK AND SECTOR SHOULD B
E READ NEXT.":.L,81,85,1:.KY. ./..L,760,
960,11:.L,81,85,0:."... THIS BYTE IS VER
Y IMPORTANT. YOU.M.0.." SHOULD NOT CHANG
E IT BECAUSE IT TELLS.{.1.." THE DRIVE T
HAT DOS1 IS TO BE USED. ANY...2.." ATTEM
PT TO ALTER IT COULD LEAD TO A...3.." DO
S MISMATCH ERROR BEING GENERATED.":.L,85
,87,1:.KY...4..L,85,87,0:."..... THESE A
RE THE 140 BYTES KNOWN AS THE".L.5.." BA
M. IT IS HARD TO BELIEVE THAT SUCH A.{.6
.." SMALL NUMBER OF BYTES COULD HOLD ALL
"...7.." THE INFORMATION REGARDING THE
STATUS"...8.." OF EACH OF THE 683 BLOCK
S ON THE DISK.":.L,89,440,1:.KY...9..L,8
9,440,0:.L,760,960,11.6.:.."..... THIS I
S THE ASCII REPRESENTATION OF".M.;.." TH
E DISK HEADER.":.L,440,480,1:.KY:.L,440,
480,0...<..".. THE IDENTIFICATION CODE I
S REPRESENTED.I.=.." BY THESE FIVE BYTES
, AGAIN THE ASCII...>.." CODES ARE USED.
":.L,485,495,1:.KY:.L,485,495,0.1.?.."..
. THE NEXT STAGE OF THIS DEMONSTRATION
".].@.." DEALS WITH ONE TRACK ONLY. USIN
G THE...A.." CURSOR KEYS, PLEASE SELECT
ANY TRACK...B.." ON THE DISK TO BE ANALY
SED. PRESS...C.." ANY OTHER KEY TO CONTI
NUE.":.2023,32:T.1...D..". TRACK:
".;.E..L,81..(T.4).40.(T .3).8,89.
.(T.4).40.(T .3).8,1:.198,0.L.F..".".7)T
". ".^.G..A$:.A$."".71...H..A$..".".A$..
".".A$..".".A$..".".79...I.T1.T:.A$.".".
T.T.4.(T.32):.77.D.J..A$.".".T.T.4.(T.4)
:.77...K..A$.".".T.T.1.(T.35):.77...L..A
$.".".T.T.1.(T.1).0.M..L,81..(T1.4).40.(
T1.3).8,89..(T1.4).40.(T1.3).8,0.8.N..69
.C.O..".. THE FOUR BYTES BELOW ARE THE B
AM"...P.." REPRESENTATION OF TRACK"T".:.
".C.Q..". ";:.I.0.3:.15,"B-P:8"T.4.I:.#8
,A$:B(I).F(A$.G(0))...R..H$(.(B(I).16))H
$((B(I).16..(B(I).16)).16);" .";:..2.S
..:."..... REMEMBER, THE FIRST BYTE IN T
HE SET OF"._.T..". FOUR IS A CHECKSUM AN
D IT TELLS THE"...U..". DRIVE HOW MANY F
REE SECTORS THERE ARE."...V..". IF THERE
IS A DISCREPANCY BETWEEN THIS"...W..".
FIGURE AND THE NUMBER OF SECTORS FREE"..
.X..". ACCORDING TO THE OTHER THREE BYTE
S A".H.Y..". DIRECTORY ERROR WILL BE GEN
ERATED.".Y.Z..". FOR NOW THOUGH, WE SHAL
L IGNORE IT.":.KY...[..I.0.3:.L,170.I.40
,200.I.40,1:..U.\..L,360,960,11:."......
.....................":.I.1.8...]..".":.
J.1.4:." "G(20):.J,I.$.^.."........ THES
E THREE BYTES STORE ALL THE".Q._..". NEC
ESSARY INFORMATION ABOUT ALL THE"...`.."
. SECTORS IN THIS TRACK.":."... BEFORE Y
OU SEE EXACTLY HOW THE INFO IS.I.A..". I
NTERPRETED INTO SOMETHING WE CAN"...B.."
. UNDERSTAND, IT IS BEST IF THE VALUES".
;.C..". ARE CONVERTED INTO BINARY....":.
KY:.L,240,960,11:.L,0,200,0.|.D.."....."
:.I.0.2:..I.9.1);:B$."":B.B(I.1):.J.0.7:
F.B..(B.2).2...E.B$."1".B$:.F.0.B$."0".J
(B$,2).B.F.B..(B.2):.:B$(I).B$:.B$" ";:.
:....G.."... PERHAPS NOW IT IS POSSIBLE
TO SEE JUST"...H..". HOW THE RECORDS ARE
KEPT BUT THERE.M.I..". IS STILL ONE ALT
ERATION THAT NEEDS TO.Z.J..". BE MADE. T
HE BITS SHOULD BE ARRANGED...K..". IN AS
CENDING ORDER OF VALUE. THUS WE.T.L..".
NEED TO CREATE A MIRROR IMAGE OF THE...M
..". BITS IN EACH OF THE THREE BYTES....
.....":.KY:.L,240,960,11.Y.N..K.0.2:A$.B
$(K):B$."":.I.0.8:D$.I(A$,1):C$.D$:A$.H(
" ".A$,8)...O..J.7.I.0..1:..K.9.1)A$".."
D$"........."B$"..."...P.B2$(K).H(B$,8).
U.Q.B$.H(B$." ",I.J).C$."
":D$." ":.J,I...R.....S..I.0.30:.1224.I,
B(1304.I):.1304.I,32:..9.T.."..... OK. N
OW LET'S STRING TOGETHER ALL THE".I.U.."
. DIGITS TO FORM ONE 24 DIGIT NUMBER. WE
"...V..". CAN ALSO FORGET ABOUT THE HEXA
DECIMAL".Z.W..". REPRESENTATION OF THE B
YTES.":B$.B2$(0).B2$(1).B2$(2):.KY...X..
L,0,40,11:.L,80,960,11./.Y..49152,0,200,
11:.49152,240,960,11:.".......";:.I.1.4.
L.Z.." ..":."
"B$".":.J.1.50:.J,I...{..".... IT MIGHT
AS WELL BE REVEALED NOW THAT".M.|..". T
HE COMPUTER LOOKS AT EACH BIT (DIGIT)...
}..". IN TURN AND EACH ONE REPRESENTS ON
E".*.~..". SECTOR ON THE DISK. HOWEVER,
YOU CHOSE".W...". TRACK"T"TO BE ANALYSE
D AND THIS HAS".V....". ONLY";:S.21:.T.1
7.S.19......T.24.S.18......T.30.S.17....
..S"SECTORS AND SO WE NEED TO":.". REMOV
E SOME EXCESS DIGITS...":.KY.. ...L,80,9
60,11:.K.1.24.S:.J.0.3:.55361.K,C(J):.M.
1.50:.M,J,K.L ..."....... NOW WE HAVE TH
E CORRECT NUMBER OF".| ...". DIGITS TO R
EPRESENT THE SECTORS ON THE".. ...". CHO
SEN TRACK.".E ..."... IF I REPLACE ALL T
HE ZEROS WITH THE ".. ...". LETTER 'X'
AND ALL THE ONES WITH A"..!...". FULL S
TOP THEN IT IS SIMPLE TO SEE".K!...". WH
ICH SECTORS ARE FREE AND WHICH ARE"..!..
.". USED.":.KY:.L,80,960,11:.I.1.S:C.46:
.B(1064.I).48.C.88.D!...J.0.3:.55336.I,C
(J):.M.1.20:.M,J:.1064.I,C:.J.3.0..1..!.
..55336.I,C(J):.T.1.20:.T,J,I.."..."....
.. OK. NOW YOU SEE THE BREAKDOWN OF THE"
.I"...". SECTORS USED ON TRACK"T".. THIS
HAS":.". BEEN OBTAINED FROM JUST THREE"
;.."..." BYTES.":.". EMPTY SECTORS ARE D
ENOTED BY THE FULL".Q"...". STOPS - ORIG
INALLY THE BITS THAT WERE.."...". SET TO
ONE. SECTOR 0 IS ON THE LEFT".$#...". A
ND SECTOR"S.1"ON THE RIGHT.".P#...".. LE
T'S RECAP QUICKLY THE STEPS FOR".|#...".
FINDING OUT WHICH SECTORS OF WHICH"..#.
..". BLOCKS ARE FREE...":.KY.H#..."...NO
TE: THIS METHOD IS FOR MANUALLY"..#..."
CHECKING THE BAM. IF DOING IT"."$.
.." FROM BASIC THEN SKIP THIS PAGE
.".R$...". 1.. WRITE DOWN THE BINARY CON
VERSION OF..$..." THE THREE BYTES.":.
".. 2.. WRITE DOWN THE MIRROR IMAGES OF"
..$..." THESE BYTES.":.".. 3.. STRING
THE THREE SETS OF NEW DIGITS"..%..."
TOGETHER TO FORM A 24 DIGIT NUMBER.".?%
...".. 4.. ERASE EXCESS DIGITS - YOU ONL
Y NEED".I% .." THE SECTORS THAT EXIST
IN THAT"..%..." TRACK TO BE REPRESEN
TED. CHECK YOUR".C%..." MANUAL IF YOU
ARE UNSURE OF THE"..%..." NUMBER OF
SECTORS IN EACH TRACK."..&...".. 5.. NOW
, EVERY DIGIT THAT IS A ONE".I&..." R
EPRESENTS A FREE SECTOR AND EACH".W&..."
ZERO REPRESENTS A USED SECTOR.":.KY.
.&...".. BASIC PROGRAM FOR READING THE B
AM:".V&...".. 10 OPEN15,8,15:OPEN8,8,8,
"G(34)"#"G(34)..'..." 20 PRINT#15,"G(34
)"U1 8 0 18 0"G(34).0'..." 30 INPUT"G(3
4)"WHICH TRACK"G(34)";TR.^'..." 40 PRIN
T#15,"G(34)"B-P:8"G(34)"TR*4+1..'..." 5
0 FOR BY=0TO2:GET#8,A$:":." B(BY)=AS
C(A$+CHR$(0))..'..." 60 NEXT BY: CLOSE8
:CLOSE15":." 70 S=20:IF TR>17 THEN S=18
".2(..." 80 IF TR>24 THEN S=17":." 90
IF TR>30 THEN S=16":." 100 SR=0: DIM SS(
S).G(..." 110 FOR BY=0TO2:B=B(BY)":." 12
0 FOR BIT=0TO7"..(..." 130 V= INT(B/2):
SS=B-(V*2)":." 140 IF SR<=S THEN SS(SR)=
1-SS".S(..." 150 B=V: SR=SR+1: NEXT BIT,
BY"..)..." .THE STATUS OF EACH SECTOR IS
HELD IN".-)..." AN ARRAY SS(X) WHERE 0=
FREE,1=USED.".\)...".. THE ABOVE PROGRAM
WILL REMAIN IN THE"..)..." MEMORY WHEN
THE DEMO HAS FINISHED.":.KY..)..."......
.I SHALL NOW READ THE BAM OF THIS DISK".
.)...". USING A SIMILAR METHOD TO THE AB
OVE"..*...". PROGRAM. TO ENABLE IT TO FI
T ON THE".E*...". DISPLAY THE HORIZONTAL
AXIS IS THE".U*...". TRACKS AND THE VER
TICAL IS THE SECTORS"..*..."... USED SEC
TORS ARE DENOTED BY SQUARE"..*...". BLOC
KS (. .).":.KY..*...".. 1
2 3":." ";:.I.1.35.9+..
.I(D(I),1);:.:.". ...................
..................Y+...I.0.20:.". "I(D(
I),1)"..".O+...I.10.I.20..".."I..+A..:..
34)"...........":..28)"...........":.I.1
.2.A+B..,," ..................":...+C..T
.1.35:.15,"B-P:8"T.4.1:.Y.0.2:.#8,A$:B(Y
).F(A$.G(0))..,D..:S.20:.T.17.S.18.!,E..
T.24.S.17.0,F..T.30.S.16.R,G..781,3:.782
,3.T:.783,0:.65520..,H.R.0:.Y.0.2:B.B(Y)
:.BT.0.7:V..(B.2):SS.B.V.2..,I..R.S.204.
.,J.."..V..";:.SS.0..".P..";:.204..,K.."
..";..,L.B.V:R.R.1:.BT,Y,T: 8: 15:.KY..
-M..".... IF YOU SHOULD EVER WISH TO CHA
NGE THE".<-N.." BAM MANUALLY THEN FOLLOW
THE SIMPLE".K-O.." STEPS BELOW (YOU'LL
HAVE TO WORK OUT A"..-P.." PROGRAM FOR T
HIS ONE - I CAN'T PROVIDE".C-Q.." EVERYT
HING FOR YOU, NOW CAN I ?)"...R..".. .1.
.READ TRACK 18, SECTOR 0":.". .2. .GET
THE APPROPRIATE THREE BYTES.?.S.." TH
AT REPRESENT THE DESIRED TRACK".O.T.."..
3. .TO FIND WHICH OF THE THREE BYTES:".
..U.." B=INT(SECTOR/8) .[RANGE 0-2]"
.C.V..".. 4. .TO FIND WHICH BIT OF BYTE
B:"...W.." BT=SECTOR-INT(SECTOR/8)*8
.[0-7]".9/X..".. 5. .TO MAKE THE SECTOR
FREE:":." BYTEVALUE=BYTEVALUE OR 2^B
T../Y..". .TO MAKE THE SECTOR USED:":
." BYTEVALUE=BYTEVALUE AND (255-2^BT)
../Z..KY:.".. DEMONSTRATION OVER. PRESS
ANY KEY TO"../..." RETURN TO BASIC WITH
THE PROGRAM TO"..0..." READ THE BAM IN M
EMORY...":.KY:."....NEW".T0..."..10OPEN1
5,8,15:OPEN8,8,8,"G(34)"#"G(34)":PRINT#1
5,";..0...G(34)"U1 8 0 18 0"G(34)":INPUT
"G(34)"WHICH TRACK"G(34)";TR.W0..."20 PR
INT#15,"G(34)"B-P:8"G(34)"TR*4+1:FORBY=0
TO2:GET#8,A$:";."1..."B(BY)=ASC(A$+CHR$(
0)):NEXT:CLOSE8":."30 CLOSE15:S=20:IFTR>
17THENS=18.R1..."40 IFTR>24THENS=17":."5
0 IFTR>30THENS=16"..1..."60 SR=0:DIMSS(S
):FORBY=0TO2:B=B(BY):FORBT=0TO7:V=INT(B/
2):SS=B-(V*2).Z1..."70 IFSR<=STHENSS(SR)
=1-SS":."80 B=V:SR=SR+1:NEXTBT,BY"..2...
"?"G(34)"...."G(34)";.";:.I.0.9:.631.I,1
3:.:.198,10.X2...".....................
......... ",".. .... . .... . ........ "
..2..."......... ......... .........","
......... ....... ......... "..2..." ...
........ ............. . ","...........
........ ......... ".)3...".............
..... ......... ","......... ... .... .
.... . ".L3..."................. ......
... ","................. ......... "..3.
..".. .............. ...... .","........
........... ......... "..3..."..........
..... ......... ",".... ..... ........
........ ".74..."............. ..... ..
.......","............. ..... ..... ".
..
> CLICK IMAGE PREVIEW FOR FULL MODAL