M.C.TUTOR 2
FILE INFORMATION
FILENAME(S): M.C.TUTOR 2
FILE TYPE(S): PRG
FILE SIZE: 26.3K
FIRST SEEN: 2025-11-30 21:08:22
APPEARS ON: 2 disk(s)
FILE HASH
087ff22069c0c33f72dea3a1c916a302e133b26bb209bcc2eec50c68c3560759
FOUND ON DISKS (2 DISKS)
| DISK TITLE | FILENAME | FILE TYPE | COLLECTION | TRACK | SECTOR | ACTIONS |
|---|---|---|---|---|---|---|
| TAPE PRG SNELL | M.C.TUTOR 2 | PRG | DuncanTwain | 26 | 0 | DOWNLOAD FILE |
| CCD 3001 | M.C.TUTOR 2 | PRG | DuncanTwain | 10 | 2 | DOWNLOAD FILE |
FILE CONTENT & ANALYSIS
> DETECTED: COMMODORE BASIC PROGRAM (LOAD ADDRESS $0801)
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000011F0: 3A 43 30 02 0D 0A 51 20 10 0D 0A 58 20 2D 20 52 |:C0...Q ...X - R|
00001200: 45 47 49 53 54 45 52 20 3A 43 30 02 0D 0A 51 20 |EGISTER :C0...Q |
00001210: 10 0D 0A 59 20 2D 20 52 45 47 49 53 54 45 52 20 |...Y - REGISTER |
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00001230: 20 46 4C 41 47 43 20 20 23 3A 30 20 20 0D 0A 44 | FLAGC #:0 ..D|
00001240: 45 43 49 4D 41 4C 20 4D 4F 44 45 20 3A 30 20 20 |ECIMAL MODE :0 |
00001250: 0D 0A 49 4E 54 45 52 52 55 50 54 53 43 20 0F 3A |..INTERRUPTSC .:|
00001260: 30 20 20 0D 0A 5A 45 52 4F 20 46 4C 41 47 44 20 |0 ..ZERO FLAGD |
00001270: 20 2C 3A 30 20 20 0D 0A 4E 45 47 41 54 49 56 45 | ,:0 ..NEGATIVE|
00001280: 20 46 4C 41 47 3A 30 20 20 0D 0A 4F 56 45 52 46 | FLAG:0 ..OVERF|
00001290: 4C 4F 57 20 46 4C 41 47 3A 30 20 20 0D 0A 51 20 |LOW FLAG:0 ..Q |
000012A0: 20 34 0D 0A 81 66 33 2F 66 37 20 46 4F 52 20 4D | 4...f3/f7 FOR M|
000012B0: 4F 44 45 53 0D 0A 84 27 52 45 54 55 52 4E 27 20 |ODES...'RETURN' |
000012C0: 54 4F 20 52 55 4E 20 20 0D 0A 82 27 5F 27 20 54 |TO RUN ...'_' T|
000012D0: 4F 20 45 44 49 54 44 20 20 37 0D 0A FF 1F 11 00 |O EDITD 7......|
000012E0: 82 9D 87 4C 41 42 45 4C 53 20 49 4E 53 54 52 55 |...LABELS INSTRU|
000012F0: 43 54 49 4F 4E 53 0D 0A 9C 0D 0A 50 52 45 53 53 |CTIONS.....PRESS|
00001300: 20 4B 45 59 53 3A 9C 0D 0A 9C 0A 81 9D 87 20 66 | KEYS:........ f|
00001310: 31 43 20 1F 9C 0D 0A 61 73 73 65 6D 62 6C 65 73 |1C ....assembles|
00001320: 20 70 72 6F 67 72 61 6D 9C 0D 0A 9C 0A 81 9D 87 | program........|
00001330: 20 66 33 43 20 1B 9C 0D 0A 66 6F 72 20 6D 61 69 | f3C ....for mai|
00001340: 6E 20 6D 65 6E 75 9C 0D 0A 9C 0A 81 9D 87 20 66 |n menu........ f|
00001350: 35 43 20 15 9C 0D 0A 74 6F 20 65 72 61 73 65 20 |5C ....to erase |
00001360: 70 72 6F 67 72 61 6D 9C 0D 82 0A 9C 0A 9C 1C 0A |program.........|
00001370: 9C 1F 00 12 20 84 9D 87 45 44 49 54 20 4D 4F 44 |.... ...EDIT MOD|
00001380: 45 20 20 9C 8C 0D 0A 20 84 9D 87 49 2A 20 4F 20 |E .... ...I* O |
00001390: 20 9C 8C 0D 0A 0A 54 79 70 65 20 69 6E 20 70 72 | .....Type in pr|
000013A0: 6F 67 72 61 6D 2C 20 75 73 69 6E 67 20 74 68 65 |ogram, using the|
000013B0: 20 63 75 72 73 6F 72 20 6B 65 79 73 0D 0A 74 6F | cursor keys..to|
000013C0: 20 65 64 69 74 20 70 72 6F 67 72 61 6D 2E 9C 0D | edit program...|
000013D0: 0A 9C FF 41 43 43 55 4D 55 4C 41 54 4F 52 44 20 |...ACCUMULATORD |
000013E0: 0A 58 2D 52 45 47 49 53 54 45 52 44 20 0F 59 2D |.X-REGISTERD .Y-|
000013F0: 52 45 47 49 53 54 45 52 20 20 81 9D 87 43 30 04 |REGISTER ...C0.|
00001400: 20 20 87 9C 45 20 03 84 9D 87 43 30 04 20 20 87 | ..E ....C0. .|
00001410: 9C 44 20 03 82 9D 87 43 30 04 20 20 87 9C 48 20 |.D ....C0. ..H |
00001420: 20 29 4E 45 47 20 5A 45 52 4F 20 44 45 43 20 4F | )NEG ZERO DEC O|
00001430: 2F 46 20 43 41 52 52 59 20 49 2F 44 49 53 41 42 |/F CARRY I/DISAB|
00001440: 4C 45 46 4C 41 47 53 3A 85 9D 87 82 30 20 20 20 |LEFLAGS:....0 |
00001450: 82 20 30 20 20 03 20 20 30 48 20 28 EC 30 FF 45 |. 0 . 0H (.0.E|
00001460: 52 52 4F 52 20 2D 20 49 4E 53 54 52 55 43 54 49 |RROR - INSTRUCTI|
00001470: 4F 4E 20 4E 4F 54 20 52 45 43 4F 47 4E 49 53 45 |ON NOT RECOGNISE|
00001480: 44 20 3A 0D 45 52 52 4F 52 20 2D 20 4E 55 4D 42 |D :.ERROR - NUMB|
00001490: 45 52 20 42 49 47 47 45 52 20 54 48 41 4E 20 32 |ER BIGGER THAN 2|
000014A0: 35 35 3A 0D 45 52 52 4F 52 20 2D 20 4C 41 42 45 |55:.ERROR - LABE|
000014B0: 4C 20 4E 4F 54 20 46 4F 55 4E 44 20 3A 20 0D 45 |L NOT FOUND : .E|
000014C0: 52 52 4F 52 20 2D 20 52 45 53 55 4C 54 49 4E 47 |RROR - RESULTING|
000014D0: 20 41 44 44 52 45 53 53 20 47 52 45 41 54 45 52 | ADDRESS GREATER|
000014E0: 20 54 48 41 4E 0D 0A 31 30 32 34 20 28 44 45 43 | THAN..1024 (DEC|
000014F0: 49 4D 41 4C 29 20 4F 52 20 24 34 30 30 20 28 48 |IMAL) OR $400 (H|
00001500: 45 58 29 FF 42 52 4B 00 43 4C 43 18 43 4C 44 D8 |EX).BRK.CLC.CLD.|
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00001520: 49 4E 58 E8 49 4E 59 C8 4E 4F 50 EA 50 48 41 48 |INX.INY.NOP.PHAH|
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00001550: 54 41 59 A8 54 53 58 BA 54 58 41 8A 54 58 53 9A |TAY.TSX.TXA.TXS.|
00001560: 54 59 41 98 FF 41 53 4C 0A 4C 53 52 4A 52 4F 4C |TYA..ASL.LSRJROL|
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00001580: 50 C9 43 50 58 E0 43 50 59 C0 45 4F 52 49 4C 44 |P.CPX.CPY.EORILD|
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000015A0: 43 E9 FF 41 44 43 65 41 4E 44 25 41 53 4C 06 42 |C..ADCeAND%ASL.B|
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000015C0: 45 43 C6 45 4F 52 45 49 4E 43 E6 4C 44 41 A5 4C |EC.EOREINC.LDA.L|
000015D0: 44 58 A6 4C 44 59 A4 4C 53 52 46 4F 52 41 05 52 |DX.LDY.LSRFORA.R|
000015E0: 4F 4C 26 52 4F 52 66 53 42 43 E5 53 54 41 85 53 |OL&RORfSBC.STA.S|
000015F0: 54 58 86 53 54 59 84 FF 41 44 43 6D 41 4E 44 2D |TX.STY..ADCmAND-|
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00001610: 43 50 59 CC 45 4F 52 4D 49 4E 43 EE 4C 44 41 AD |CPY.EORMINC.LDA.|
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00001630: 52 4F 4C 2E 52 4F 52 6E 53 42 43 ED 53 54 41 8D |ROL.RORnSBC.STA.|
00001640: 53 54 58 8E 53 54 59 8C 44 45 43 CE FF 41 44 43 |STX.STY.DEC..ADC|
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00001670: BC 4C 53 52 5E 4F 52 41 1D 52 4F 4C 3E 52 4F 52 |.LSR^ORA.ROL>ROR|
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000016D0: 44 59 B4 4C 53 52 56 4F 52 41 15 52 4F 4C 36 52 |DY.LSRVORA.ROL6R|
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00001750: 42 4D 49 30 42 4E 45 D0 42 50 4C 10 42 56 43 50 |BMI0BNE.BPL.BVCP|
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000017D0: 11 C8 B1 14 85 12 C8 B1 14 85 13 20 7D 23 C8 B1 |........... }#..|
000017E0: 14 85 12 C8 B1 14 85 13 20 7D 23 C8 B1 14 85 12 |........ }#.....|
000017F0: C8 B1 14 85 13 20 7D 23 C8 B1 14 8D 5B C0 20 A7 |..... }#....[. .|
00001800: 23 A2 00 BD 07 27 C9 FF F0 08 CD 5B C0 F0 0F E8 |#....'.....[....|
00001810: D0 F1 A5 19 C9 0C 90 03 4C E1 26 20 57 28 60 6C |........L.& W(`l|
00001820: 10 00 8C 02 C0 A0 00 B1 12 C9 FF F0 06 20 10 34 |............. .4|
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00001940: 83 24 C9 0C D0 1D AE 5D C0 BD 01 7B 8D 02 C0 A0 |.$.....]...{....|
00001950: 01 B1 0E CD 02 C0 F0 08 C8 C0 11 D0 F4 4C 20 28 |.............L (|
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00001970: 18 69 02 8D 02 C0 18 AC 09 C0 71 0E 8D 02 C0 A0 |.i........q.....|
00001980: 01 4C AC 24 60 A0 00 AD 06 C0 29 6C 8D 06 C0 A0 |.L.$`.....)l....|
00001990: 00 B1 18 8D 5E C0 0E 5E C0 08 AD 5E C0 91 18 68 |....^..^...^...h|
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000019C0: 06 C0 60 AD 07 C0 AE 0F C0 9D 00 09 CE 0F C0 60 |..`............`|
000019D0: EE 0F C0 AE 0F C0 BD 00 09 8D 07 C0 4C B5 26 20 |............L.& |
000019E0: 25 34 68 68 4C 4E 0C A0 00 B1 18 8D 07 C0 4C B5 |%4hhLN........L.|
000019F0: 26 AD 07 C0 A0 00 91 18 60 A0 00 B1 18 8D 04 C0 |&.......`.......|
00001A00: 4C B5 26 A0 00 AD 04 C0 91 18 60 A0 00 B1 18 8D |L.&.......`.....|
00001A10: 05 C0 4C B5 26 A0 00 AD 05 C0 91 18 60 AE 5D C0 |..L.&.......`.].|
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000026F0: 8D 6A C0 A9 20 4C 61 32 C9 5F D0 05 A9 1F 4C 61 |.j.. La2._....La|
00002700: 32 C9 61 90 02 E9 60 8D 63 C0 AD 66 C0 AC 65 C0 |2.a...`.c..f..e.|
00002710: 91 22 A9 00 85 26 85 27 AE 63 C0 E0 00 F0 11 18 |."...&.'.c......|
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00002750: 20 4C E9 32 18 A5 24 69 40 85 29 85 24 A5 25 69 | L.2..$i@.).$.%i|
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000027E0: 91 20 A9 20 91 22 C8 D0 F5 E6 21 E6 23 A5 21 C9 |. . ."....!.#.!.|
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00002800: 21 A9 00 8D 65 C0 A9 A0 85 2A A9 00 85 24 A9 A0 |!...e....*...$..|
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000028C0: C0 C9 0D F0 03 4C AE 31 4C AE 32 8A 48 98 48 20 |.....L.1L.2.H.H |
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000028E0: C0 0D 67 C0 0D 69 C0 F0 05 A9 00 8D 66 C0 A0 3C |..g..i......f..<|
000028F0: 88 F0 08 A2 FF CA D0 FD 4C 4A 34 68 A8 68 AA AD |........LJ4h.h..|
00002900: 66 C0 60 48 A9 00 8D 68 C0 8D 67 C0 8D 69 C0 68 |f.`H...h..g..i.h|
00002910: C9 85 D0 06 A9 01 8D 68 C0 60 C9 86 D0 06 A9 01 |.......h.`......|
00002920: 8D 69 C0 60 C9 87 D0 05 A9 01 8D 67 C0 60 AC 65 |.i.`.......g.`.e|
00002930: C0 B1 20 29 F0 0D 6B C0 91 20 C8 C0 28 D0 F2 A9 |.. )..k.. ..(...|
00002940: 20 4C 61 32 C9 87 F0 01 60 AC 65 C0 B1 22 AA A9 | La2....`.e.."..|
00002950: 87 60 02 05 07 06 04 03 01 08 09 0A 0B 0C 0D 0E |.`..............|
00002960: 0F A9 00 85 76 A9 8C 85 77 A2 12 A0 14 B1 76 29 |....v...w.....v)|
00002970: 0F C9 04 F0 18 C8 C0 28 D0 F3 18 A5 76 69 28 85 |.......(....vi(.|
00002980: 76 A5 77 69 00 85 77 CA D0 E3 4C EE 34 A9 10 91 |v.wi..w...L.4...|
00002990: 76 4C CF 34 AC 65 C0 B1 20 29 F0 09 04 91 20 60 |vL.4.e.. ).... `|
000029A0: AC 5A C0 AD 13 C0 91 0E EE 5A C0 60 A9 1F 20 A6 |.Z.......Z.`.. .|
000029B0: 31 A9 00 20 A6 31 20 A6 31 A2 28 A9 20 20 A6 31 |1.. .1 .1.(. .1|
000029C0: CA 10 F8 A9 01 8D 02 C0 A9 1F 20 A6 31 A9 00 20 |.......... .1.. |
000029D0: A6 31 AD 02 C0 20 A6 31 A2 13 A9 20 20 A6 31 CA |.1... .1... .1.|
000029E0: 10 F8 EE 02 C0 AD 02 C0 C9 12 D0 DC A9 1F 20 A6 |.............. .|
000029F0: 31 A9 00 20 A6 31 A9 11 20 A6 31 A2 C8 A9 20 20 |1.. .1.. .1... |
00002A00: A6 31 CA D0 F8 20 2D 1A 60 A9 77 85 12 A9 35 85 |.1... -.`.w...5.|
00002A10: 13 20 C0 22 20 57 10 20 25 34 4C 4E 0C 1F 00 12 |. ." W. %4LN....|
00002A20: 45 4E 44 20 4F 46 20 50 52 4F 47 52 41 4D 20 2D |END OF PROGRAM -|
00002A30: 20 50 52 45 53 53 20 41 20 4B 45 59 FF 61 63 63 | PRESS A KEY.acc|
00002A40: 75 6D 75 6C 61 74 6F 72 20 FF 58 20 72 65 67 69 |umulator .X regi|
00002A50: 73 74 65 72 20 FF 59 20 72 65 67 69 73 74 65 72 |ster .Y register|
00002A60: 20 FF 4C 6F 67 69 63 61 6C 20 4F 52 20 FF 77 69 | .Logical OR .wi|
00002A70: 74 68 20 6E 75 6D 62 65 72 20 FF 77 69 74 68 20 |th number .with |
00002A80: 6D 65 6D 6F 72 79 20 6C 6F 63 61 74 69 6F 6E FF |memory location.|
00002A90: 41 72 69 74 68 6D 65 74 69 63 20 73 68 69 66 74 |Arithmetic shift|
00002AA0: 20 6C 65 66 74 20 FF 50 75 73 68 20 FF 70 72 6F | left .Push .pro|
00002AB0: 63 65 73 73 6F 72 20 73 74 61 74 75 73 20 72 65 |cessor status re|
00002AC0: 67 69 73 74 65 72 0D FF 6F 6E 74 6F 20 73 74 61 |gister..onto sta|
00002AD0: 63 6B 20 FF 6D 65 6D 6F 72 79 20 6C 6F 63 61 74 |ck .memory locat|
00002AE0: 69 6F 6E 20 FF 42 72 65 61 6B 20 2D 20 77 69 6C |ion .Break - wil|
00002AF0: 6C 20 73 74 6F 70 20 65 78 65 63 75 74 69 6F 6E |l stop execution|
00002B00: FF 4E 6F 74 20 61 20 76 61 6C 69 64 20 69 6E 73 |.Not a valid ins|
00002B10: 74 72 75 63 74 69 6F 6E 20 6F 70 2D 63 6F 64 65 |truction op-code|
00002B20: FF 0D 50 52 45 53 53 20 52 45 54 55 52 4E 20 4B |..PRESS RETURN K|
00002B30: 45 59 20 54 4F 20 43 4F 4E 54 49 4E 55 45 FF 63 |EY TO CONTINUE.c|
00002B40: 6C 65 61 72 20 FF 63 61 72 72 79 20 FF 66 6C 61 |lear .carry .fla|
00002B50: 67 20 FF 72 69 67 68 74 20 FF 6C 65 66 74 20 FF |g .right .left .|
00002B60: 52 6F 74 61 74 65 20 FF 4C 6F 67 69 63 61 6C 20 |Rotate .Logical |
00002B70: 41 4E 44 20 61 63 63 75 6D 75 6C 61 74 6F 72 20 |AND accumulator |
00002B80: FF 54 65 73 74 20 42 49 54 73 20 6F 66 20 61 63 |.Test BITs of ac|
00002B90: 63 75 6D 75 6C 61 74 6F 72 20 61 67 61 69 6E 73 |cumulator agains|
00002BA0: 74 20 FF 66 72 6F 6D 20 73 74 61 63 6B FF 50 75 |t .from stack.Pu|
00002BB0: 6C 6C 20 FF 4C 6F 67 69 63 61 6C 20 73 68 69 66 |ll .Logical shif|
00002BC0: 74 20 FF 45 78 63 6C 75 73 69 76 65 20 4F 52 20 |t .Exclusive OR |
00002BD0: FF 73 65 74 20 FF 77 69 74 68 20 6D 65 6D 6F 72 |.set .with memor|
00002BE0: 79 0D 6C 6F 63 61 74 69 6F 6E FF 41 64 64 20 77 |y.location.Add w|
00002BF0: 69 74 68 20 FF 69 6E 74 65 72 72 75 70 74 20 64 |ith .interrupt d|
00002C00: 69 73 61 62 6C 65 20 FF 54 72 61 6E 73 66 65 72 |isable .Transfer|
00002C10: 20 FF 53 74 6F 72 65 20 FF 61 74 20 6D 65 6D 6F | .Store .at memo|
00002C20: 72 79 20 6C 6F 63 61 74 69 6F 6E FF 74 6F 20 73 |ry location.to s|
00002C30: 74 61 63 6B 20 70 6F 69 6E 74 65 72 FF 74 6F 20 |tack pointer.to |
00002C40: 61 63 63 75 6D 75 6C 61 74 6F 72 FF 4A 75 6D 70 |accumulator.Jump|
00002C50: 20 74 6F 20 73 75 62 72 6F 75 74 69 6E 65 FF 4A | to subroutine.J|
00002C60: 75 6D 70 20 2D 20 4C 6F 61 64 20 70 72 6F 67 72 |ump - Load progr|
00002C70: 61 6D 20 63 6F 75 6E 74 65 72 20 77 69 74 68 20 |am counter with |
00002C80: 6E 65 77 0D 61 64 64 72 65 73 73 FF 44 65 63 72 |new.address.Decr|
00002C90: 65 6D 65 6E 74 20 FF 4C 6F 61 64 20 FF 73 74 61 |ement .Load .sta|
00002CA0: 63 6B 20 70 6F 69 6E 74 65 72 20 FF 74 6F 20 58 |ck pointer .to X|
00002CB0: 20 72 65 67 69 73 74 65 72 FF 6F 76 65 72 66 6C | register.overfl|
00002CC0: 6F 77 20 FF 4E 6F 20 6F 70 65 72 61 74 69 6F 6E |ow .No operation|
00002CD0: FF 49 6E 63 72 65 6D 65 6E 74 20 FF 64 65 63 69 |.Increment .deci|
00002CE0: 6D 61 6C 20 6D 6F 64 65 20 FF 6E 75 6D 62 65 72 |mal mode .number|
00002CF0: 20 FF 53 75 62 74 72 61 63 74 20 77 69 74 68 20 | .Subtract with |
00002D00: FF 43 6F 6D 70 61 72 65 20 FF 6E 6F 74 20 65 71 |.Compare .not eq|
00002D10: 75 61 6C 20 74 6F 20 7A 65 72 6F 20 28 5A 3D 30 |ual to zero (Z=0|
00002D20: 29 FF 65 71 75 61 6C 20 74 6F 20 7A 65 72 6F 20 |).equal to zero |
00002D30: 28 5A 3D 31 29 FF 42 72 61 6E 63 68 20 69 66 20 |(Z=1).Branch if |
00002D40: FF 70 6F 73 69 74 69 76 65 20 28 4E 3D 30 29 FF |.positive (N=0).|
00002D50: 6D 69 6E 75 73 20 28 4E 3D 31 29 FF 52 65 74 75 |minus (N=1).Retu|
00002D60: 72 6E 20 66 72 6F 6D 20 73 75 62 72 6F 75 74 69 |rn from subrouti|
00002D70: 6E 65 FF 74 6F 20 59 20 72 65 67 69 73 74 65 72 |ne.to Y register|
00002D80: FF 73 65 74 20 28 43 3D 31 29 FF 63 6C 65 61 72 |.set (C=1).clear|
00002D90: 20 28 43 3D 30 29 FF 6F 76 65 72 66 6C 6F 77 20 | (C=0).overflow |
00002DA0: 66 6C 61 67 20 63 6C 65 61 72 20 28 56 3D 30 29 |flag clear (V=0)|
00002DB0: FF 6F 76 65 72 66 6C 6F 77 20 66 6C 61 67 20 73 |.overflow flag s|
00002DC0: 65 74 20 28 56 3D 31 29 FF 00 A9 01 8D 02 C0 A9 |et (V=1)........|
00002DD0: 14 8D 0B C0 A9 00 8D 14 C0 4C 4E 39 A9 1F 20 A6 |.........LN9.. .|
00002DE0: 31 AD 0B C0 20 A6 31 AD 02 C0 20 A6 31 A9 87 20 |1... .1... .1.. |
00002DF0: 9E 34 8A 60 20 36 39 AE 14 C0 9D 51 C0 EE 14 C0 |.4.` 69....Q....|
00002E00: EE 0B C0 AD 14 C0 C9 06 D0 EA AD 51 C0 C9 20 D0 |...........Q.. .|
00002E10: 03 4C AA 39 A9 01 8D 63 C0 A9 14 8D 0B C0 A9 00 |.L.9...c........|
00002E20: 8D 14 C0 AD 63 C0 CD 02 C0 F0 1B 20 00 3A AE 14 |....c...... .:..|
00002E30: C0 DD 51 C0 D0 10 EE 14 C0 EE 0B C0 AD 14 C0 C9 |..Q.............|
00002E40: 06 D0 E8 4C B8 39 EE 63 C0 AD 63 C0 C9 11 D0 C9 |...L.9.c..c.....|
00002E50: EE 02 C0 AD 02 C0 C9 11 F0 03 4C 29 39 60 A9 DD |..........L)9`..|
00002E60: 85 12 A9 39 85 13 20 57 10 A2 00 BD 51 C0 20 A6 |...9.. W....Q. .|
00002E70: 31 E8 E0 06 D0 F5 A9 22 20 A6 31 20 25 34 68 68 |1......" .1 %4hh|
00002E80: 4C 4E 0C 1F 00 12 45 52 52 4F 52 20 2D 20 44 55 |LN....ERROR - DU|
00002E90: 50 4C 49 43 41 54 45 20 4C 41 42 45 4C 20 46 4F |PLICATE LABEL FO|
00002EA0: 55 4E 44 20 22 FF A9 1F 20 A6 31 AD 0B C0 20 A6 |UND "... .1... .|
00002EB0: 31 AD 63 C0 20 A6 31 A9 87 20 9E 34 8A 60 A9 1F |1.c. .1.. .4.`..|
00002EC0: 20 A6 31 A9 1A 20 A6 31 AD 28 C0 20 A6 31 A9 87 | .1.. .1.(. .1..|
00002ED0: 20 9E 34 E0 20 F0 03 4C CC 3B A2 03 BD 15 C0 C9 | .4. ..L.;......|
00002EE0: 20 F0 1B 4C 47 3B AD 15 C0 C9 42 F0 07 C9 4A F0 | ..LG;....B...J.|
00002EF0: 03 4C A6 3A E8 BD 15 C0 C9 20 D0 F8 F0 4E E8 BD |.L.:..... ...N..|
00002F00: 15 C0 E0 05 F0 08 C9 20 F0 42 C9 41 B0 D8 C9 24 |....... .B.A...$|
00002F10: F0 13 E0 05 F0 08 C9 28 F0 41 C9 23 F0 E0 C9 30 |.......(.A.#...0|
00002F20: B0 06 4C 1D 3B 4C 22 3B C9 3A B0 F6 E8 E0 0D F0 |..L.;L";.:......|
00002F30: F1 BD 15 C0 C9 20 F0 14 C9 2C F0 0F C9 30 B0 03 |..... ...,...0..|
00002F40: 4C 1D 3B C9 3A B0 DB 4C 86 3A 60 E8 E8 E0 0D F0 |L.;.:..L.:`.....|
00002F50: F9 BD 15 C0 C9 20 F0 F4 4C 1D 3B E8 BD 15 C0 C9 |..... ..L.;.....|
00002F60: 29 F0 0B C9 28 F0 BB E0 0C D0 F0 4C 1D 3B E8 BD |)...(......L.;..|
00002F70: 15 C0 C9 29 F0 AC C9 28 F0 A8 E0 0C D0 F0 A2 05 |...)...(........|
00002F80: BD 15 C0 C9 24 F0 1D BD 15 C0 C9 29 F0 0F C9 2C |....$......)...,|
00002F90: F0 0B C9 30 90 0B C9 3A B0 07 E8 D0 EA E8 4C A5 |...0...:......L.|
00002FA0: 3A 4C 1D 3B E8 BD 15 C0 C9 29 F0 F1 C9 2C F0 ED |:L.;.....)...,..|
00002FB0: C9 30 90 ED C9 47 B0 E9 C9 41 B0 E8 C9 3A B0 E1 |.0...G...A...:..|
00002FC0: 4C FE 3A 68 68 4C 9D 14 E8 BD 15 C0 C9 20 F0 17 |L.:hhL....... ..|
00002FD0: C9 2C F0 16 C9 30 90 C9 C9 3A 90 EC C9 41 90 C1 |.,...0...:...A..|
00002FE0: C9 47 B0 BD 4C 22 3B 4C A6 3A 4C A5 3A A9 55 85 |.G..L";L.:L.:.U.|
00002FF0: 12 A9 3B 85 13 20 57 10 4C D7 3B 1F 00 12 54 48 |..;.. W.L.;...TH|
00003000: 45 52 45 20 53 48 4F 55 4C 44 20 42 45 20 41 20 |ERE SHOULD BE A |
00003010: 53 50 41 43 45 20 42 45 54 57 45 45 4E 20 54 48 |SPACE BETWEEN TH|
00003020: 45 0D 0A 4D 4E 45 4D 4F 4E 49 43 20 41 4E 44 20 |E..MNEMONIC AND |
00003030: 4F 50 45 52 41 4E 44 2E 20 54 48 45 20 48 49 47 |OPERAND. THE HIG|
00003040: 48 4C 49 47 48 54 45 44 0D 0A 4C 49 4E 45 20 49 |HLIGHTED..LINE I|
00003050: 4E 44 49 43 41 54 45 53 20 57 48 49 43 48 20 4C |NDICATES WHICH L|
00003060: 49 4E 45 20 48 41 53 20 41 4E 20 45 52 52 4F 52 |INE HAS AN ERROR|
00003070: 2E FF A9 03 85 12 A9 3C 85 13 20 57 10 A9 1F 20 |.......<.. W... |
00003080: A6 31 A9 10 20 A6 31 AD 28 C0 20 A6 31 A9 88 20 |.1.. .1.(. .1.. |
00003090: A6 31 A9 81 20 A6 31 A9 9D 20 A6 31 A9 87 20 A6 |.1.. .1.. .1.. .|
000030A0: 31 20 25 34 68 68 4C 4E 0C 1F 00 12 41 20 4C 41 |1 %4hhLN....A LA|
000030B0: 42 45 4C 20 53 48 4F 55 4C 44 20 4F 4E 4C 59 20 |BEL SHOULD ONLY |
000030C0: 42 45 20 55 50 54 4F 20 53 49 58 20 4C 45 54 54 |BE UPTO SIX LETT|
000030D0: 45 52 53 20 4C 4F 4E 47 2E 20 54 48 45 20 48 49 |ERS LONG. THE HI|
000030E0: 47 48 4C 49 47 48 54 45 44 20 4C 49 4E 45 20 49 |GHLIGHTED LINE I|
000030F0: 4E 44 49 43 41 54 45 53 20 54 48 45 4C 49 4E 45 |NDICATES THELINE|
00003100: 20 57 48 49 43 48 20 48 41 53 20 41 20 4C 41 42 | WHICH HAS A LAB|
00003110: 45 4C 20 54 4F 4F 20 4C 4F 4E 47 2E FF 0B 8E 3C |EL TOO LONG....<|
00003120: AB 3C CD 3C F0 3C 17 3D 3A 3D 5D 3D 80 3D A9 3D |.<.<.<.=:=]=.=.=|
00003130: CC 3D F3 3D 01 43 20 1A 4C 65 73 73 6F 6E 20 38 |.=.=.C .Lesson 8|
00003140: 20 2D 20 54 72 61 6E 73 66 65 72 73 0D 0A FF 16 | - Transfers....|
00003150: 3E 02 45 20 1D 45 78 65 72 63 69 73 65 20 31 20 |>.E .Exercise 1 |
00003160: 66 6F 72 20 6C 65 73 73 6F 6E 20 38 0D 0A FF CC |for lesson 8....|
00003170: 3F 01 43 20 20 20 4C 65 73 73 6F 6E 20 39 20 2D |?.C Lesson 9 -|
00003180: 20 42 69 6E 61 72 79 20 6E 6F 74 61 74 69 6F 6E | Binary notation|
00003190: 0D 0A FF 55 40 01 43 20 20 24 4C 65 73 73 6F 6E |...U@.C $Lesson|
000031A0: 20 31 30 20 2D 20 53 68 69 66 74 73 20 61 6E 64 | 10 - Shifts and|
000031B0: 20 72 6F 74 61 74 65 73 0D 0A FF B3 4A 02 45 20 | rotates....J.E |
000031C0: 1E 45 78 65 72 63 69 73 65 20 31 20 66 6F 72 20 |.Exercise 1 for |
000031D0: 6C 65 73 73 6F 6E 20 31 30 0D 0A FF 9B 53 01 43 |lesson 10....S.C|
000031E0: 20 20 20 4C 65 73 73 6F 6E 20 31 31 20 2D 20 54 | Lesson 11 - T|
000031F0: 77 6F 20 6D 6F 72 65 20 66 6C 61 67 73 0D 0A FF |wo more flags...|
00003200: 70 54 02 45 20 1E 45 78 65 72 63 69 73 65 20 31 |pT.E .Exercise 1|
00003210: 20 66 6F 72 20 6C 65 73 73 6F 6E 20 31 31 0D 0A | for lesson 11..|
00003220: FF F7 58 01 43 20 20 26 4C 65 73 73 6F 6E 20 31 |..X.C &Lesson 1|
00003230: 32 20 2D 20 4C 6F 67 69 63 61 6C 20 69 6E 73 74 |2 - Logical inst|
00003240: 72 75 63 74 69 6F 6E 73 0D 0A FF AA 59 02 45 20 |ructions....Y.E |
00003250: 1E 45 78 65 72 63 69 73 65 20 31 20 66 6F 72 20 |.Exercise 1 for |
00003260: 6C 65 73 73 6F 6E 20 31 32 0D 0A FF E9 64 01 43 |lesson 12....d.C|
00003270: 20 20 24 4C 65 73 73 6F 6E 20 31 33 20 2D 20 49 | $Lesson 13 - I|
00003280: 6E 64 65 78 65 64 20 61 64 64 72 65 73 73 69 6E |ndexed addressin|
00003290: 67 0D 0A FF A6 65 02 45 20 21 87 45 78 65 72 63 |g....e.E !.Exerc|
000032A0: 69 73 65 20 31 20 66 6F 72 20 6C 65 73 73 6F 6E |ise 1 for lesson|
000032B0: 20 31 33 0D 0A FF 1F 6D 1F 10 02 54 52 41 4E 53 | 13....m...TRANS|
000032C0: 46 45 52 53 0D 0A 0A 20 54 68 65 72 65 20 61 72 |FERS... There ar|
000032D0: 65 20 71 75 69 74 65 20 61 20 66 65 77 20 74 72 |e quite a few tr|
000032E0: 61 6E 73 66 65 72 0D 0A 69 6E 73 74 72 75 63 74 |ansfer..instruct|
000032F0: 69 6F 6E 73 20 6F 6E 20 74 68 65 20 36 35 30 32 |ions on the 6502|
00003300: 2E 20 41 20 74 72 61 6E 73 66 65 72 20 69 73 20 |. A transfer is |
00003310: 74 68 65 20 63 6F 70 79 69 6E 67 20 6F 66 20 6F |the copying of o|
00003320: 6E 65 20 76 61 6C 75 65 20 6F 66 20 61 20 72 65 |ne value of a re|
00003330: 67 69 73 74 65 72 20 20 69 6E 74 6F 20 61 6E 6F |gister into ano|
00003340: 74 68 65 72 20 72 65 67 69 73 74 65 72 2E 0D 0A |ther register...|
00003350: 0A 54 68 65 20 66 6F 6C 6C 6F 77 69 6E 67 20 74 |.The following t|
00003360: 72 61 6E 73 66 65 72 73 20 61 72 65 20 61 76 61 |ransfers are ava|
00003370: 69 6C 61 62 6C 65 0D 0A 0A 81 54 41 58 87 2D 54 |ilable....TAX.-T|
00003380: 52 41 4E 53 46 45 52 20 41 43 43 55 4D 55 4C 41 |RANSFER ACCUMULA|
00003390: 54 4F 52 20 54 4F 20 58 2D 52 45 47 49 53 54 45 |TOR TO X-REGISTE|
000033A0: 52 0A 82 54 58 41 87 2D 54 52 41 4E 53 46 45 52 |R..TXA.-TRANSFER|
000033B0: 20 58 2D 52 45 47 49 53 54 45 52 20 54 4F 20 41 | X-REGISTER TO A|
000033C0: 43 43 55 4D 55 4C 41 54 4F 52 0A 83 54 41 59 87 |CCUMULATOR..TAY.|
000033D0: 2D 54 52 41 4E 53 46 45 52 20 41 43 43 55 4D 55 |-TRANSFER ACCUMU|
000033E0: 4C 41 54 4F 52 20 54 4F 20 59 2D 52 45 47 49 53 |LATOR TO Y-REGIS|
000033F0: 54 45 52 0A 84 54 59 41 87 2D 54 52 41 4E 53 46 |TER..TYA.-TRANSF|
00003400: 45 52 20 59 2D 52 45 47 49 53 54 45 52 20 54 4F |ER Y-REGISTER TO|
00003410: 20 41 43 43 55 4D 55 4C 41 54 4F 52 0A 0A 20 54 | ACCUMULATOR.. T|
00003420: 68 65 82 72 65 20 61 20 44 6C 73 6F 20 74 77 6F |he.re a Dlso two|
00003430: 20 6F 74 68 65 72 20 74 72 61 6E 73 66 65 72 0D | other transfer.|
00003440: 0A 69 6E 73 74 72 75 63 74 69 6F 6E 73 20 74 68 |.instructions th|
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00004CC0: 43 20 23 36 37 0D 47 20 04 53 45 43 0D 47 20 04 |C #67.G .SEC.G .|
00004CD0: 43 4C 56 0D 47 20 09 53 42 43 20 23 31 32 30 0D |CLV.G .SBC #120.|
00004CE0: 47 20 04 43 4C 56 0D 47 20 04 43 4C 43 0D 47 20 |G .CLV.G .CLC.G |
00004CF0: 07 41 44 43 20 23 31 0D 47 20 22 0E 42 52 4B FF |.ADC #1.G ".BRK.|
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00004D70: 69 6E 73 74 72 75 63 74 69 6F 6E 20 72 65 6C 69 |instruction reli|
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7.5S7.Q&B7.5.7.:%[6{6.(.:%[6{6.(.L%L7.5S
7.:%[6{6.(.:%[6{6.(.F%.7.5H5.B%.7.5U5.T%
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6.(..'.8 6.8.B%.7.5U5.:%[6{6.(.:%[6{6.(.
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8 6D8..'.8.(.(.:%[6{6.(._'[8.5U5.;'L8 6.
6..'+8.6.(.:%[6{6.(.U'.8|8.(.;'L8 6.6.:%
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6.(..F.H.H.H..)...X.D...LW3.F.I.P....K..
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I(. .!I..!.."I(.".#I..#.$I@.$.%I..%L.3ND
....K....J....P. .4L.3.F.H.H.H.F.I...L.1
L.2.H.H .. ..X.. ]4.F..H..G..I......F. <
.....JP.LJ4H.H..F.`H...H..G..I.HI.P....H
.`I.P....I.`I.P....G.`.E.. )..K.. H.(P..
LA2I...`.E.."...`..................V...
W.. ..V).I...H.(P...VI(.V.WI..WJP.L.4...
VLO4.E.. ).... `.Z.......Z.`.. .1.. .1 .
1.(. .1J......... .1.. .1... .1... .1J
........I.P... .1.. .1.. .1.H. .1JP. -.
`.W...5.. ." W. %4LN....END OF PROGRAM -
PRESS A KEY.ACCUMULATOR .X REGISTER .Y
REGISTER .LOGICAL OR .WITH NUMBER .WITH
MEMORY LOCATION.ARITHMETIC SHIFT LEFT .P
USH .PROCESSOR STATUS REGISTER..ONTO STA
CK .MEMORY LOCATION .BREAK - WILL STOP E
XECUTION.NOT A VALID INSTRUCTION OP-CODE
..PRESS RETURN KEY TO CONTINUE.CLEAR .CA
RRY .FLAG .RIGHT .LEFT .ROTATE .LOGICAL
AND ACCUMULATOR .TEST BITS OF ACCUMULATO
R AGAINST .FROM STACK.PULL .LOGICAL SHIF
T .EXCLUSIVE OR .SET .WITH MEMORY.LOCATI
ON.ADD WITH .INTERRUPT DISABLE .TRANSFER
.STORE .AT MEMORY LOCATION.TO STACK POI
NTER.TO ACCUMULATOR.JUMP TO SUBROUTINE.J
UMP - LOAD PROGRAM COUNTER WITH NEW.ADDR
ESS.DECREMENT .LOAD .STACK POINTER .TO X
REGISTER.OVERFLOW .NO OPERATION.INCREME
NT .DECIMAL MODE .NUMBER .SUBTRACT WITH
.COMPARE .NOT EQUAL TO ZERO (Z=0).EQUAL
TO ZERO (Z=1).BRANCH IF .POSITIVE (N=0).
MINUS (N=1).RETURN FROM SUBROUTINE.TO Y
REGISTER.SET (C=1).CLEAR (C=0).OVERFLOW
FLAG CLEAR (V=0).OVERFLOW FLAG SET (V=1)
.................LN9.. .1... .1... .1..
.4.` 69....Q..........I.P..Q.I P.L.9...C
............C.M.... .:....Q.P..........I
.P.L.9.C..C.I.PI......I...L)9`.....9.. W
....Q. .1...P.." .1 %4HHLN....ERROR - DU
PLICATE LABEL FOUND "... .1... .1.C. .1.
. .4.`.. .1.. .1.(. .1.. .4. ..LL;.....I
..LG;...IB..IJ..L.:....I P..N........I
.BIA.XI$......I(.AI#..I0..L.;L";I:......
....I ..I,..I0..L.;I:..L.:`.........I ..
L.;....I)..I(....P.L.;....I)..I(....P...
...I$.....I)..I,..I0..I:...P..L.:L.;....
I)..I,..I0..IG..IA..I:..L.:HHL......I ..
I,..I0.II:..IA.AIG..L";L.:L.:.U...;.. W.
LW;...THERE SHOULD BE A SPACE BETWEEN TH
E..MNEMONIC AND OPERAND. THE HIGHLIGHTED
..LINE INDICATES WHICH LINE HAS AN ERROR
.......<.. W... .1.. .1.(. .1.. .1.. .1.
. .1.. .1 %4HHLN....A LABEL SHOULD ONLY
BE UPTO SIX LETTERS LONG. THE HIGHLIGHTE
D LINE INDICATES THELINE WHICH HAS A LAB
EL TOO LONG....<.<M<.<.=:=]=.=.=L=.=.C .
LESSON 8 - TRANSFERS....>.E .EXERCISE 1
FOR LESSON 8...L?.C LESSON 9 - BINARY
NOTATION...U@.C $LESSON 10 - SHIFTS AND
ROTATES....J.E .EXERCISE 1 FOR LESSON 1
0....S.C LESSON 11 - TWO MORE FLAGS...
PT.E .EXERCISE 1 FOR LESSON 11....X.C &
LESSON 12 - LOGICAL INSTRUCTIONS....Y.E
.EXERCISE 1 FOR LESSON 12....D.C $LESSO
N 13 - INDEXED ADDRESSING....E.E !.EXERC
ISE 1 FOR LESSON 13....M...TRANSFERS...
THERE ARE QUITE A FEW TRANSFER..INSTRUCT
IONS ON THE 6502. A TRANSFER IS THE COPY
ING OF ONE VALUE OF A REGISTER INTO ANO
THER REGISTER....THE FOLLOWING TRANSFERS
ARE AVAILABLE....TAX.-TRANSFER ACCUMULA
TOR TO X-REGISTER..TXA.-TRANSFER X-REGIS
TER TO ACCUMULATOR..TAY.-TRANSFER ACCUMU
LATOR TO Y-REGISTER..TYA.-TRANSFER Y-REG
ISTER TO ACCUMULATOR.. THE.RE A DLSO TWO
OTHER TRANSFER..INSTRUCTIONS THAT WILL
BE DISCUSSED..LATER..G .LDA #104.G .TXA.
G .LDX #40.G .TXA.G .CLC.G .ADC #20.G .T
AX.G .TAY.G .LDY #3.G .TYA.G !|BRK....BI
NARY NOTATION... DECIMAL IS A SYSTEM OF
USING 10..DIFFERENT NUMBERS IN EACH NUMB
ER COLUMN.THIS SYSTEM IS CALLED BASE 10.
HOWEVER, THIS IS NOT THE SYSTEM USED BY
A..COMPUTER, WHICH USES BASE 2. THIS SY
STEMUSES ONLY TWO DIFFERENT NUMBERS FOR
EACHNUMBER COLUMN, ONE AND ZERO. BECAUSE
OF THIS, THE COLUMN HEADINGS WILL BE..D
IFFERENT. THE COLUMN HEADINGS FOR..DECIM
AL ARE:...1C0C .100C .10C [1....BINARY
NOTATION... SO THE DECIMAL NUMBER 2398 I
S ACTUALLY CALCULATED THUS:...2 TIMES 1C
0D .=C .2C0...3 TIMES 100E .=D .300..9 T
IMES 10F .=E .90..8 TIMES 1G .= +C .8..
T D- ...GIVING A RESULT OF 2398....BINA
RY NOTATION... IN BINARY THE COLUMN HEAD
INGS ARE LIKE THIS:...128 64 32 16 8
4 2 1...HERE IS AN EXAMPLE OF CONVER
TING BINARY TO DECIMAL:...1F .128..0G .+
0..0G .+0C .SO THE BINARY NUMBER..1F .+1
6C .10010C1. WOULD BE..0G .+0C .151 (128
+16+4+2+1)..1G .+4C .IN DECIMAL..1G .+2.
.1G .+1..G C-...G !(151....BINARY NOTATI
ON... EACH BINARY DIGIT (1 OR 0) IS CALL
ED..A BIT. THE PREVIOUS EXAMPLE WAS AN E
IGHTBIT NUMBER. THE BITS ARE NUMBERED FR
OM RIGHT TO LEFT AS BIT 0 TO BIT 7. ALL
..THE REGISTERS AND MEMORY LOCATIONS..ON
THE 6502 ARE EIGHT BITS IN LENGTH...SO
THE MAXIMUM VALUE THEY CAN HOLD IS..H1"S
BINARY, OR 255 DECIMAL (128+64+32+16+8+
4+2+1). A GROUP OF EIGHT BITS..IS COMMON
LY CALLED A BYTE... IN THE WAY BINARY IS
USED ABOVE A BYTE COULD ONLY BE USED TO
REPRESENT A..POSITIVE NUMBER. HOWEVER,
BY USING A..TECHNIQUE CALLED TWO'S COMPL
EMENT..NEGATIVE NUMBERS CAN ALSO BE STOR
ED.....BINARY NOTATION... WHEN USING TWO
'S COMPLEMENT TO..REPRESENT NEGATIVE NUM
BERS, BIT 7 (THE..FAR LEFT MOST BIT OF A
BYTE) REPRESENTS..THE SIGN. ON THE 6502
A ONE IS USED FOR..A NEGATIVE NUMBER AN
D A ZERO FOR A..POSITIVE NUMBER. TO CHAN
GE THE SIGN OF ANUMBER IS QUITE SIMPLE.
FIRST YOU HAVE..TO REVERSE EVERY DIGIT S
O THAT A ONE..BECOMES A ZERO AND A ZERO
BECOMES A ONE.THEN YOU HAVE TO ADD ONE F
OR THE FINAL..RESULT.....BINARY NOTATION
...C0.11010 WOULD BE:... REVERSE C0.110
10 TO GIVE..J C1.00101 AND THEN ADD ONE.
.I .+G0.1..J H-...J C1 .00110...THEREFOR
E THE NEGATIVE EQUIVALENT OF..C0.11010 I
S C1 .00110.....BINARY NOTATION... USING
TWO'S COMPLEMENT YOU CAN USE..ADDITION
TO SIMULATE SUBTRACTION. TAKE..THE EXAMP
LE OF 43-1. THE NEGATIVE OF..POSITIVE ON
E IS H1 L. AND 43 IN..BINARY IS 00101011
. THEREFORE, ADDING..THEM TOGETHER GIVES
:...C .00101011 +..C H1...C H-"R..(1)0
0101010...NOTE THAT THERE IS A CARRY, WH
ICH MUST..BE IGNORED IF USING TWO'S COMP
LEMENT. SOTHE RESULT OF THE ADDITION IS
00101010..WHICH IS 42, THE CORRECT ANSWE
R.....SHIFTS AND ROTATES... SHIFTS AND R
OTATES ARE INSTRUCTIONS..THAT DIRECTLY R
ELATE TO CHANGING OF..BINARY NUMBERS. EA
CH INSTRUCTION WILL..NOW BE DESCRIBED...
NOTE THAT THE INSTRUCTION MAY BE..FOLLO
WED BY A NUMBER REPRESENTING A..MEMORY L
OCATION, OR THE LETTER A..REPRESENTING A
CCUMULATOR. THE FUNCTION..OF THE INSTRUC
TION WILL THEN BE CARRIED..OUT ON THE ME
MORY LOCATION OR THE..ACCUMULATOR.....RO
TATE RIGHT.. ...ROR ... ...C*!B .... T
HIS INSTRUCTION PERFORMS WHAT IS KNOWNAS
A NINE BIT ROTATION. ALL THE BITS IN..T
HE BINARY NUMBER ARE ROTATED ONE PLACE..
TO THE RIGHT. THE CARRY HOLDS THE DATA..
CONTAINED IN BIT 0. THE OLD DATA OF THE
CARRY FLAG IS THEN STORED IN BIT 7...THE
FOLLOWING EXAMPLE SHOULD MAKE..THINGS C
LEARER:...BEFORE: CARRY NUMBER..J ..1.
D .0.1.0.1.1.0.1.0...AFTER:D ..0D 1.1.0
.1.0.1.1.0.1....ROTATE LEFT.. ...ROL ..
. ...C* G .... THIS INSTRUCTION IS ALSO
KNOWN AS A..NINE BIT ROTATION. IT IS TH
E SAME AS THEROR INSTRUCTION EXCEPT THE
BITS ARE..ROTATED LEFT. THE FOLLOWING EX
AMPLE..SHOULD MAKE THIS CLEAR:...BEFORE:
NUMBERL .CARRY..H ..0.1.1.1.1.1.0.1D .
.1...AFTER: .1.1.1.1.1.0.1.1D +.0....L
OGICAL SHIFT RIGHT.. ...LSR ... ...C*!G
.... THIS INSTRUCTION IS SIMILAR TO TH
E ROR INSTRUCTION, EXCEPT THAT IT IS A E
IGHT BIT ROTATION (A SHIFT). A ZERO ENT
ERS..THE LEFT END, BIT 7, AND SHIFTS ALL
THE OTHER NUMBERS ONE PLACE TO THE RIGH
T...THE DIGIT THAT FALLS OFF THE RIGHT E
ND..GOES INTO THE CARRY. THE LSR INSTRUC
TIONHAS THE EFFECT OF DIVIDING A NUMBER
BY TWO, WITH THE CARRY HOLDING A RESULT
..REPRESENTING THE HALF DIGIT. HERE IS A
N EXAMPLE DEMONSTRATING LSR:...BEFORE:
CARRY NUMBER..J ..X.C .1.1.1.0.0.1.0.0.
..AFTER:D ..0.C :0.1.1.1.0.0.1.0....ARI
THMETIC SHIFT LEFT.. ...ASL ... ...C*!X
.... THIS INSTRUCTION IS SIMILAR TO LS
R..EXCEPT THAT THE BITS ARE SHIFTED TO T
HE LEFT. THIS HAS THE EFFECT OF MULTIPLY
INGTHE NUMBER BY TWO. IF THE CARRY IS SE
T AFTER THE INSTRUCTION, THIS INDICATES
..TOO LARGE A RESULT, AND THAT 256 SHOUL
D BE ADDED TO THE RESULT TO GET THE..COR
RECT ANSWER... WITH ASL A ZERO ENTERS TH
E FAR RIGHT..END, BIT 0, AND SHIFTS ALL
THE OTHER..BITS ONE PLACE TO THE LEFT. W
HAT WAS IN BIT 7 IS PLACED IN THE CARRY
FLAG....BEFORE: CARRY NUMBER..I ..XC .
.1.0.0.0.0.1.1.0...AFTER:C ..1C ..0.0.0.
0.1.1.0.0.G .LDA #1.G .ROL A.G .ROL A.G
.STA 100.G .ROL A.G .SEC.G .ASL 100.G .L
DA 100.G .ROR A.G .LDA #128.G .ASL A.G .
CLC.G .LDA #2.G .LSR A.G .ROR A.G ".BRK.
...TWO MORE FLAGS....THE NEGATIVE FLAG..
. THIS FLAG ALWAYS REFLECTS THE STATUS O
FBIT 7 AFTER AN INSTRUCTION. IF BIT 7 IS
SET AFTER AN INSTRUCTION THEN THIS FLAG
WILL BE SET. IF BIT 7 IS CLEARED AFTER.
.AN INSTRUCTION THEN THIS FLAG WILL BE..
CLEARED.....THE OVERFLOW FLAG... THIS FL
AG IS SET IF THE NEGATIVE FLAG HAS BEEN
SET OR CLEARED BY AN ARITHMETICOPERATIO
N, WHOSE RESULT THEN HAS THE..WRONG SIGN
.....TWO MORE FLAGS...IF YOU ADD TWO LAR
GE NUMBERS TOGETHER..THEN THERE IS A CHA
NCE OF THE OVERFLOW FLAG BEING SET. TAK
E THIS EXAMPLE:-...C .01D0.11 (+67).. +
01F0. (+64).. I-...C .1E0"011 (-125)...
...TWO MORE FLAGS... IN THE EXAMPLE THE
TWO POSITIVE..NUMBERS ADDED TOGETHER HAV
E PRODUCED A..NEGATIVE RESULT. IF THIS H
APPENS THEN..THE OVERFLOW FLAG WILL BECO
ME SET. THIS FLAG WILL BECOME SET IF:-..
.ADDING LARGE POSITIVE NUMBERS..ADDING L
ARGE NEGATIVE NUMBERS..SUBTRACTING A LAR
GE POSITIVE NUMBER FROMA LARGE NEGATIVE
NUMBER..SUBTRACTING A LARGE NEGATIVE NUM
BER FROMA LARGE POSITIVE NUMBER.... NOTE
THAT THE INSTRUCTION CLV STANDS..FOR CL
EAR OVERFLOW FLAG. THERE IS A..DEMONSTRA
TION PROGRAM SHOWING WHEN THE OVERFLOW
AND NEGATIVE FLAGS ARE SET..G .SEC.G .LD
A #130.G .SBC #20.G .CLC.G .LDA #65.G .C
LV.G .ADC #67.G .SEC.G .CLV.G .SBC #120.
G .CLV.G .CLC.G .ADC #1.G ".BRK....LOGIC
AL INSTRUCTIONS... LOGICAL INSTRUCTIONS
ARE BASED ON THE MATHEMATICAL WORK OF G
EORGE BOOLE. EACH LOGIC INSTRUCTION RELI
ES ON WHAT IS..KNOWN AS A 'TRUTH TABLE'.
THE TWO INPUTSARE SHOWN AT THE TOP, WIT
H THE RELATING RESULT AT THE BOTTOM. ALL
LOGICAL..INSTRUCTIONS ARE PERFORMED BET
WEEN THE ACCUMULATOR AND A MEMORY LOCAT
ION OR AN IMMEDIATE NUMBER. EACH PAIR OF
BITS ARE DONE SEPARATELY BY THE 6502...
. EACH LOGICAL INSTRUCTION AVAILABLE TO
THE 6502 WILL NOW BE DESCRIBED.....LOGI
CAL INSTRUCTIONS... ..AND .... ..C* .
.... WITH THE AND INSTRUCTION, EACH PAIR
OF BITS OF THE ACCUMULATOR AND MEMORY..
LOCATION OR NUMBER ARE COMPARED. IF THEY
ARE BOTH ONES, THEN THE OUTPUT WILL BE
ONE, OTHERWISE THE OUTPUT WILL BE ZERO.
HERE IS THE TRUTH TABLE FOR AND:-...INPU
T 1:. 0 . 1C .1..INPUT 2: 0C .1C .0C
.1..J M-...RESULTS:. 0 ( 1...AN EXAMP
LE:..10101101 ANDED WITH..D1C0.1 GIVES:.
.101D0!91....LOGICAL INSTRUCTIONS... THE
AND INSTRUCTION IS USEFUL WHEN YOU WANT
TO 'MASK OFF' CERTAIN BITS OF A..BINARY
NUMBER. FOR EXAMPLE, SAY BITS 7 TO 4 A
RE UNIMPORTANT, THEN YOU COULD USETHE AN
D INSTRUCTION TO ERASE THE BITS..THAT AR
E NOT WANTED. THE FOLLOWING..EXAMPLE SHO
WS THIS:...11011010 - ORIGINAL NUMBER.
.D0D1. - THE 'MASK' BYTE..D0 .1010 -
THE RESULT... THE RESULT NOW HAS THE UP
PER FOUR BITS MISSING. NOTE THAT IN THE
EXAMPLE..PROGRAMS YOU WILL HAVE TO CONVE
RT FROM DECIMAL TO BINARY FOR THE RESUL
TS TO..MAKE SENSE.....LOGICAL INSTRUCTIO
NS.. ...ORA ... ...C* P .... WITH THE
INSTRUCTION ORA, WHICH STANDS FOR OR ACC
UMULATOR, EACH PAIR OF BITS..ARE COMPARE
D, IF ONE OR BOTH THE BITS..ARE SET THEN
THE RESULT WILL BE A ONE... HERE IS THE
TRUTH TABLE FOR ORA:...INPUT 1:. 0 .
1C .1..INPUT 2: 0C .1C .0C .1..J M-...R
ESULTS: 0C .1 `1....LOGICAL INSTRUCT
IONS...HERE IS AN EXAMPLE USING ORA:...1
0101011 - NUMBER TO BE ORED WITH:..D0D1.
- THIS NUMBER GIVES..1010D1 . - THIS RE
SULT.... THE ORA INSTRUCTION IS USEFUL I
F YOU..WANT TO MAKE SURE CERTAIN BITS AR
E A ONEAS IN THE PREVIOUS EXAMPLE.....LO
GICAL INSTRUCTIONS.. ...EOR ... ...C* .
.... THIS INSTRUCTION IS SIMILAR TO TH
E ORA INSTRUCTION, EXCEPT THAT IF BOTH I
NPUTS ARE ONES, THEN THE RESULT WILL BE
ZERO. THE TRUTH TABLE FOR EOR IS:...INP
UT 1:. 0 . 1C .1..INPUT 2: 0C .1C .0C
.1..J M-...RESULTS: 0C .1 !.0... EOR
ACTUALLY STANDS FOR EXCLUSIVE OR...ONE
OF ITS USES IS FOR REVERSING..SELECTED B
ITS. EACH BIT THAT YOU WANT..REVERSED HA
S TO BE IN THE NUMBER..OR THE MEMORY LOC
ATION.....LOGICAL INSTRUCTIONS...HERE IS
AN EXAMPLE OF USING THE EOR..INSTRUCTIO
N TO REVERSE EVERY BIT. IF YOUWERE TO AD
D ONE TO THE RESULT YOU WOULD HAVE THE O
PPOSITE SIGN OF THE NUMBER. IFIT HAD BEE
N POSITIVE IT WOULD NOW BE..NEGATIVE AND
VICE VERSA....0110C1 !0 - NUMBER TO BE
EXCLUSIVE ORED..H1 " - WITH THIS NUMBER
TO GIVE:..1001C0.1 - THIS NUMBER..G .LDA
#1.G .ORA #2.G .ORA #4.G .ORA #4.G .AND
#3.G .AND #2.G .AND #0.G .LDA #255.G .S
TA 1C0..G .LDA #190.G .EOR 1C0..G .EOR #
255.G '}BRK....INDEXED ADDRESSING... SO
FAR ONLY THREE ADDRESSING MODES HAVEBEEN
USED, IMMEDIATE, ABSOLUTE AND..IMPLIED.
THE IMPLIED FORM OF ADDRESSING IS THE E
ASIEST TO UNDERSTAND, IT IS JUSTA THREE
LETTER MNEMONIC, LIKE TXA. THE..ADDRESSI
NG MODE DETERMINES WHERE THE..6502 WILL
FETCH ITS DATA FROM, FOR THE..INSTRUCTIO
N ABOUT TO BE EXECUTED. WITH..THE IMMEDI
ATE FORM OF ADDRESSING, THE..6502 USES T
HE ACTUAL NUMBER SPECIFIED...SO IF THE I
NSTRUCTION WAS LDA #20, THE 6502 WOULD
LOAD THE ACCUMULATOR WITH THENUMBER 20,
NOT THE MEMORY LOCATION.....INDEXED ADDR
ESSING... WITH ABSOLUTE ADDRESSING THE N
UMBER..GIVEN AFTER THE INSTRUCTION IS A
MEMORY..LOCATION. WITH THE INSTRUCTION L
DA 20..THE 6502 WILL LOAD THE ACCUMULATO
R WITH THE NUMBER CONTAINED IN THE MEMOR
Y..LOCATION NUMBER 20, WHICH COULD BE AN
Y..NUMBER BETWEEN 0 AND 255, WHICH NEED
NOTBE THE ACTUAL NUMBER 20.....INDEXED A
DDRESSING... INDEXED ADDRESSING IS SIMIL
AR TO..ABSOLUTE ADDRESSING IN THE FACT T
HAT THE6502 FETCHES THE DATA FROM A MEMO
RY..LOCATION. THE X AND Y REGISTERS ARE.
.SOMETIMES CALLED INDEX REGISTERS,..BECA
USE OF THEIR USE WITH THIS FORM OF..ADDR
ESSING. TWO EXAMPLES OF USING..INDEXED A
DDRESSING ARE ABOUT TO BE..GIVEN, WITH A
N EXPLANATION SHOWING HOW..THEY ARE USED
, AND HOW THEY WORK.....INDEXED ADDRESSI
NG... ONE EXAMPLE OF USING INDEXED ADDRE
SSINGIS LDA 710,X AND ANOTHER EXAMPLE IS
..ORA 654,Y. IN THE FIRST EXAMPLE THE..C
ONTENTS OF THE X REGISTER ARE ADDED TO 7
10 TO GIVE A MEMORY LOCATION ADDRESS...S
O, IF THE X REGISTER HOLDS 25 WHEN THE..
PROGRAM IS RUNNING THEN THE 6502 WILL..L
OAD THE ACCUMULATOR WITH THE CONTENTS O
F MEMORY LOCATION 735 (710+25)...SIMILAR
LY, WITH THE SECOND EXAMPLE THE..Y REGIS
TER IS ADDED TO 654 TO GIVE A..MEMORY LO
CATION ADDRESS. IF THE Y..REGISTER CONTA
INS 216 WHEN THE PROGRAM IS RUNNING THE
N THE ACCUMULATOR WILL BE..ORED AGAINST
MEMORY LOCATION 870..(654+216). AN EXAMP
LE PROGRAM SHOWS..INDEXED ADDRESSING IN
USE..G .LDA #6.G .STA 605.G .CLC.G .ADC
#2.G .STA 606.G .ADC #2.G .STA 607.G .LD
X #0.G .TXA.G .LDA 605,X.G .INX.G .ADC 6
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