The Complete Machine Code Tutor (UK) (Face 1) (1985) [Original] [TAPE] [UTILITAIRE].cdt
HexaDumpAscii only
000000: 5A 58 54 61 70 65 21 1A 01 0C 20 21 46 11 13 09 ZXTape!... !F...
000010: A6 04 57 04 94 04 27 09 01 10 08 0F 00 07 01 00 ..W...'.........
000020: 2C 4D 2F 43 20 54 55 54 4F 52 00 00 00 00 00 00 ,M/C TUTOR......
000030: 00 01 FF 02 6F 05 FA 0F FF 6F 05 00 10 00 00 00 ....o....o......
000040: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000050: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000060: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000070: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000080: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000090: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0000A0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0000B0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0000C0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0000D0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0000E0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0000F0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000100: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000110: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000120: 00 E6 79 FF FF FF FF 11 14 09 A6 04 A6 04 93 04 ..y.............
000130: 26 09 FF 0F 08 31 21 11 06 00 16 CD A1 BC D4 00 &....1!.........
000140: 00 CD FF BB 3E 00 DD 21 A3 10 DD 46 00 48 F5 CD ....>..!£..F.H..
000150: 32 BC F1 DD 23 3C FE 04 20 F0 01 0E 0E CD 38 BC 2...#<.. .....8.
000160: 21 00 C0 11 01 C0 01 FF 3F 36 00 ED B0 3E 05 CD !.......?6...>..
000170: 6F BB 3E 16 CD 72 BB 06 1F 21 A7 10 7E C5 E5 CD o.>..r...!..~...
000180: 5D BB E1 C1 23 10 F5 DD 21 C6 10 21 00 C0 11 00 ]...#...!..!....
000190: 08 CD 63 10 21 00 E0 CD 63 10 21 00 10 11 00 43 ..c.!...c.!....C
0001A0: 3E 6C 18 97 DD 7E 00 07 DD AE 00 E6 22 DD 7E 00 >l...~......".~.
0001B0: DD 23 0E 01 20 05 DD 4E 00 DD 23 B9 20 03 FE 00 .#.. ..N..#. ...
0001C0: C8 06 04 E5 F5 77 CB 67 20 06 CB 27 CB A7 18 04 .....w.g ..'....
0001D0: CB 27 CB E7 CB 4F 28 01 3C 19 10 E9 F1 E1 23 0D .'...O(.<.....#.
0001E0: 20 DF 18 C0 0E 18 09 01 A4 20 4E 65 77 20 47 65 ........ New Ge
0001F0: 6E 65 72 61 74 69 6F 6E 20 53 6F 66 74 77 61 72 neration Softwar
000200: 65 20 20 31 39 38 35 00 C0 11 F0 05 11 F0 01 00 e 1985.........
000210: 01 11 F0 01 11 F0 04 00 42 FF 01 F0 01 00 02 FF ........B.......
000220: 01 F0 04 FF 01 F0 03 00 43 FF 01 F0 01 00 02 FF ........C.......
000230: 01 F0 01 00 01 FF 01 F0 01 FF 01 0C 76 F0 04 00 ............v...
000240: D2 11 F0 01 0F 01 1F F0 01 00 01 11 F0 01 0F 01 ................
000250: 1F F0 01 FF 01 F0 02 11 F0 02 FF 01 F0 01 0F 01 ................
000260: 1F F0 01 FF 01 F0 01 00 03 FF 01 F0 01 0F 02 0E ................
000270: EF 0F 01 1F F0 01 0F 01 0E FF 01 F0 01 0F 02 0E ................
000280: 00 25 FF 01 F0 01 00 01 11 F0 01 00 01 FF 01 F0 .%..............
000290: 01 00 01 FF 01 F0 01 FF 01 F0 01 EF 0E FF 01 F0 ................
0002A0: 01 FF 01 F0 01 0F 01 0E 00 01 FF 01 F0 01 00 03 ................
0002B0: FF 01 F0 01 0F 01 0E 00 03 FF 01 F0 01 00 02 FF ................
0002C0: 01 F0 01 0F 01 0E 00 27 EF 0F 01 0E 00 03 EF 0F .......'........
0002D0: 01 0E 00 01 EF 0E 00 02 EF 0E EF 0E 00 03 EF 0F ................
0002E0: 03 0E EF 0F 03 0E 00 02 EF 0E 00 02 EF 0F 03 0E ................
0002F0: 00 69 11 F0 01 00 02 11 F0 01 00 03 11 F0 01 00 .i..............
000300: 04 11 F0 02 00 02 11 F0 01 00 01 11 F0 01 00 01 ................
000310: 11 F0 01 00 01 11 F0 01 00 02 11 F0 01 11 F0 04 ................
000320: 00 09 11 F0 02 00 02 11 F0 02 00 01 11 F0 03 00 ................
000330: 01 11 F0 04 00 0C FF 01 F0 01 1F F0 01 FF 56 1F ..............V.
000340: F0 01 00 01 11 F0 01 0E EF 1F F0 01 00 01 FF 01 ................
000350: F0 01 00 01 EF 0E 00 01 FF 01 F0 04 00 01 FF 01 ................
000360: F0 01 00 01 FF 01 F0 01 1F F0 01 FF 01 F0 01 FF ................
000370: 01 F0 03 00 09 FF 01 F0 01 00 01 EF 0E FF 01 F0 ................
000380: 01 00 01 FF 01 F0 01 FF 01 F0 01 00 01 FF 01 F0 ................
000390: 01 FF 01 F0 03 00 0D FF 01 F0 01 00 02 FF 01 F0 ................
0003A0: 01 00 01 FF 01 F0 01 0F 02 1F F0 01 00 01 EF 1F ................
0003B0: F0 02 0E 00 01 FF 01 F0 01 00 01 FF 01 F0 01 00 ................
0003C0: 01 FF 01 F0 01 00 01 FF 01 F0 01 00 01 EF 1F F0 ................
0003D0: 01 FF 01 F0 04 00 08 EF 1F F0 02 0E EF 1F F0 02 ................
0003E0: 0E FF 01 F0 03 0E FF 01 F0 04 00 C0 EF 0F 01 1F ................
0003F0: F0 01 0F 01 0E FF 01 F0 01 00 01 FF 01 F0 01 EF ................
000400: 0F 01 1F F0 01 0F 01 0E 11 F0 01 0F 01 1F F0 01 ................
000410: FF 01 F0 01 0F 01 1F F0 01 00 37 FF 01 F0 01 00 ..........7.....
000420: 02 FF 01 F0 01 00 01 FF 01 F0 01 00 02 FF 01 F0 ................
000430: 01 00 02 FF 01 F0 01 00 01 FF 01 F0 01 FF 01 DE ................
000440: A2 F0 01 1F F0 01 00 38 EF 0E 00 03 EF 0F 01 0E .......8........
000450: 00 03 EF 0E 00 03 EF 0F 01 0E 00 01 EF 0E 00 01 ................
000460: EF 0E 00 00 00 C0 EF 0F 01 1F F0 01 0F 01 0E FF ................
000470: 01 F0 01 00 01 FF 01 F0 01 FF 01 F0 01 0F 02 0E ................
000480: 00 42 FF 01 F0 01 00 02 FF 01 F0 01 0F 01 1F F0 .B..............
000490: 01 FF 01 F0 01 0F 01 0E 00 43 EF 0E 00 02 EF 0E .........C......
0004A0: 00 01 EF 0E EF 0F 03 0E 00 83 11 F0 02 00 03 11 ................
0004B0: F0 02 00 01 11 F0 01 00 02 11 F0 01 11 F0 03 00 ................
0004C0: 01 11 F0 01 00 03 11 F0 04 11 F0 05 11 F0 04 00 ................
0004D0: 25 FF 01 F0 01 00 01 EF 0E 00 01 FF 01 F0 01 00 %...............
0004E0: 01 FF 01 F0 01 FF 01 F0 01 1F F0 01 1F F0 01 FF ................
0004F0: 01 F0 03 0E FF 01 F0 01 00 03 FF 01 F0 03 00 03 ................
000500: FF 01 F0 01 00 02 FF 01 F0 03 00 26 EF 1F F0 02 ...........&....
000510: 0E 00 01 EF 1F F0 02 0E FF 01 F0 01 00 02 FF 01 ................
000520: F0 01 FF 01 F0 01 00 03 FF 01 F0 04 FF 01 F0 04 ................
000530: 00 02 FF 01 F0 01 00 02 FF 01 F0 04 00 B9 FF 01 ................
000540: F0 00 32 02 11 F0 02 00 02 11 F0 01 1F F0 01 00 ..2.............
000550: 02 11 F0 01 0F 01 1F F0 01 00 01 FF 01 F0 01 00 ................
000560: 01 FF 01 F0 01 00 01 FF 01 F0 01 00 01 FF 01 F0 ................
000570: 02 00 01 FF 01 F0 01 FF 01 F0 01 0F 02 0E 00 08 ................
000580: 11 F0 01 0F 01 1F F0 01 11 F0 01 0F 01 1F F0 01 ................
000590: FF 01 F0 01 0F 01 1F F0 01 FF 01 F0 01 0F 02 0E ................
0005A0: 00 0C FF 01 F0 01 EF 0E FF 01 F0 01 00 01 FF 01 ................
0005B0: F0 05 00 01 FF 01 F0 01 00 01 11 F0 01 00 01 FF ................
0005C0: 01 F0 01 0F 01 1F F0 01 00 01 FF 01 F0 01 00 01 ................
0005D0: FF 01 F0 01 EF 1F F0 02 FF 01 F0 01 0F 01 0E 00 ................
0005E0: 09 FF 01 F0 01 00 01 11 F0 01 FF 01 F0 01 00 01 ................
0005F0: FF 01 F0 01 FF 01 F0 01 00 01 FF 01 F0 01 FF 01 ................
000600: F0 01 0F 01 0E 00 0D EF 0E 00 02 EF 0E 00 01 EF ................
000610: 0E 00 02 EF 0E 00 02 EF 0F 01 0E 00 02 EF 0E 00 ................
000620: 01 EF 0E 00 01 EF 0E 00 01 EF 0E 00 02 EF 0E EF ................
000630: 0F 03 0E 00 09 EF 0F 01 0E 00 02 EF 0F 01 0E 00 ................
000640: 01 EF 0F 70 8F 02 0E 00 01 EF 0F 03 0E 00 70 11 ...p..........p.
000650: F0 05 11 F0 01 00 01 11 F0 01 11 F0 05 00 01 11 ................
000660: F0 02 00 01 11 F0 03 00 38 FF 01 F0 01 00 02 FF ........8.......
000670: 01 F0 01 00 01 FF 01 F0 01 00 02 FF 01 F0 01 00 ................
000680: 02 FF 01 F0 01 00 01 FF 01 F0 01 FF 01 F0 03 0E ................
000690: 00 37 FF 01 F0 01 00 02 EF 1F F0 02 0E 00 02 FF .7..............
0006A0: 01 F0 01 00 02 EF 1F F0 02 0E FF 01 F0 01 EF 1F ................
0006B0: F0 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0006C0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0006D0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0006E0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
0006F0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000700: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000710: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000720: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000730: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000740: 00 00 00 00 00 50 B3 FF FF FF FF 11 15 09 A6 04 .....P..........
000750: A6 04 91 04 23 09 1E 10 08 D8 22 8B 43 00 6C 20 ....#.....".C.l
000760: 71 BC 21 BA 10 CD 88 10 CD D9 51 38 18 CD 9B 10 q.!.......Q8....
000770: CD CB 51 CD B3 19 CD 1F 10 CD CB 51 18 E4 ED 73 ..Q........Q...s
000780: 59 14 C3 48 12 21 00 11 CD 88 10 21 B7 11 11 01 Y..H.!.....!....
000790: 00 3E FC CD A1 BC 30 3C 3A B7 11 FE 00 21 22 11 .>....0<:....!".
0007A0: 28 15 FE 02 21 40 11 28 0E FE 04 21 5F 11 28 07 (...!@.(...!_.(.
0007B0: FE 06 21 7E 11 20 24 CD 88 10 21 00 58 11 00 55 ..!~. $...!.X..U
0007C0: 3E FB CD A1 BC 30 0D CD 9B 10 3E 0F 32 91 1A CD >....0....>.2...
0007D0: E8 50 18 B1 21 E0 10 FE 00 28 03 21 9D 11 CD 88 .P..!....(.!....
0007E0: 10 CD 93 10 C3 03 10 3E F0 32 91 1A 11 00 18 C3 .......>.2......
0007F0: B0 51 06 00 CD 19 BD 10 FB C9 CD FF BB 3E 00 DD .Q...........>..
000800: 21 92 1A DD 46 00 48 F5 CD 32 BC F1 DD 23 3C FE !...F.H..2...#<.
000810: 04 20 F0 01 0E 0E C3 38 BC 20 20 44 6F 20 79 6F . .....8. Do yo
000820: 75 20 77 69 73 68 20 74 6F 20 65 6E 74 65 72 20 u wish to enter
000830: 6C 65 73 73 6F 6E 73 3F 20 20 20 59 2F 4E A3 20 lessons? Y/N£
000840: 20 20 20 20 20 20 4C 6F 61 64 69 6E 67 20 68 61 Loading ha
000850: 73 20 62 65 65 6E 20 73 74 6F 70 70 65 64 A3 34 s been stopped£4
000860: 45 20 20 20 20 20 20 20 50 6C 65 61 73 65 20 73 E Please s
000870: 74 61 72 74 20 74 61 70 65 20 72 65 63 6F 72 64 tart tape record
000880: 65 72 A3 20 20 20 20 20 20 20 20 20 20 4C 6F 61 er£ Loa
000890: 64 69 6E 67 20 6C 65 73 73 6F 6E 73 20 31 2D 39 ding lessons 1-9
0008A0: A3 20 20 20 20 20 20 20 20 20 4C 6F 61 64 69 6E £ Loadin
0008B0: 67 20 6C 65 73 73 6F 6E 73 20 31 30 2D 31 37 A3 g lessons 10-17£
0008C0: 20 20 20 20 20 20 20 20 20 4C 6F 61 64 69 6E 67 Loading
0008D0: 20 6C 65 73 73 6F 6E 73 20 31 38 2D 32 35 A3 20 lessons 18-25£
0008E0: 20 20 20 20 20 20 20 20 4C 6F 61 64 69 6E 67 20 Loading
0008F0: 6C 65 73 73 6F 6E 73 20 32 36 2D 33 35 A3 20 20 lessons 26-35£
000900: 20 20 20 20 20 20 20 20 20 20 4C 6F 61 64 69 6E Loadin
000910: 67 20 65 72 72 6F 72 A3 00 00 ED 73 59 14 CD 55 g error£...sY..U
000920: 13 CD E6 12 CD 00 14 CD DF 33 CD CB 13 21 3C 16 .........3...!<.
000930: 11 00 12 CD F1 32 C3 6A 12 CD 06 BB CD 19 BD FE .....2.j........
000940: 45 28 66 FE 54 CA 0A 13 FE 52 28 11 3E 42 CD 1E E(f.T....R(.>B..
000950: BB 28 E6 ED 7B 59 14 3E 0F 32 91 1A C9 CD E6 12 .(..{Y.>.2......
000960: CD 95 07 76 12 DD 2A EC 3B 11 00 12 ED 53 D0 31 ...v..*.;....S.1
000970: DD 7E 03 FE 72 38 08 21 EC 15 CD F1 32 18 BC CD .~..r8.!....2...
000980: 46 20 DD 7E 03 FE 72 30 B2 CD 06 BB FE 53 28 0F F .~..r0.....S(.
000990: FE 45 28 17 FE 23 28 BD CD 35 3E 30 C5 18 9C 21 .E(..#(..5>0...!
0009A0: 8C 16 11 00 12 CD F1 32 C3 D8 11 CD 8F 13 CD E6 .......2........
0009B0: 12 CD B6 18 CD 00 14 CD E6 12 ED 73 0E 3B CD DF ...........s.;..
0009C0: 33 CD CB 13 21 4B 15 11 00 12 CD F1 32 CD F3 3C 3...!K......2..<
0009D0: CD AE 3D CD E6 12 C3 D8 11 2A 12 14 CD 1A BC 0E ..=......*......
0009E0: 08 06 50 7E E6 0F 77 23 10 F9 11 B0 07 19 0D 20 ..P~..w#.......
0009F0: F0 DD 21 20 3B ED 5B CD 31 21 02 00 06 10 C5 DD ..! ;.[.1!......
000A00: CB 00 7E 28 29 DD 7E 08 BB 20 23 DD 7E 09 BA 20 ..~().~.. #.~..
000A10: 1D C1 DD 22 EC 3B 22 12 14 CD 1A BC 0E 08 06 50 ...".;"........P
000A20: 7E ED 6F 23 10 FA 11 B0 07 19 0D 20 F1 C9 01 0C ~.o#....... ....
000A30: 00 DD 09 2C C1 10 C7 3E 0F 32 91 1A 11 00 12 21 ...,...>.2.....!
000A40: 9C 15 CD F1 32 E1 C3 D8 11 DD 21 20 3B DD 22 EC ....2.....! ;.".
000A50: 3B DD 6E 08 DD 66 09 22 CD 31 21 01 02 22 D0 31 ;.n..f.".1!..".1
000A60: 21 45 17 95 3F 22 EE 3B 21 FE 0F 22 9B 42 C9 3A !E..?".;!..".B.:
000A70: CF 31 EE 01 32 CF 31 CD 21 13 CD E6 12 CD 00 14 .1..2.1.!.......
000A80: CD DF 33 C3 6A 12 CD B3 19 DD 21 20 3B 11 45 17 ..3.j.....! ;.E.
000A90: 06 10 21 CF 31 CB D6 C5 D5 DD 22 EC 3B CD D2 31 ..!.1.....".;..1
000AA0: 11 0C 00 DD 19 D1 21 17 00 19 EB C1 10 E9 21 CF ......!.......!.
000AB0: 31 CB 96 CD CB 13 CD D9 19 C9 CD B3 19 2A 41 17 1............*A.
000AC0: 11 45 17 06 10 3E 20 C5 BE 01 06 00 28 05 ED B0 .E...> .....(...
000AD0: 13 18 05 EB 0C 09 EB 23 01 10 00 CB 7E 20 0C ED .......#....~ ..
000AE0: A0 EA 76 13 C1 10 E0 CD D9 19 C9 23 ED A0 E2 7F ..v........#....
000AF0: 13 2B 18 F8 3E F0 32 91 1A 11 00 00 21 83 14 CD .+..>.2.....!...
000B00: F1 32 3E 0F 32 91 1A 3E 20 CD 45 1A 3E 14 BA 20 .2>.2..> .E.>..
000B10: F6 DD 21 3B 14 06 0A C5 DD 7E 00 32 91 1A DD 6E ..!;.....~.2...n
000B20: 01 DD 66 02 CD F1 32 01 03 00 DD 09 C1 10 E8 C9 ..f...2.........
000B30: 3E F0 32 91 1A 11 00 00 21 5B 14 CD F1 32 3E 0F >.2.....![...2>.
000B40: 32 91 1A 11 00 14 DD 21 14 14 06 0D C5 DD 7E 00 2......!......~.
000B50: 32 91 1A DD 6E 01 DD 66 02 CD F1 32 01 03 00 DD 2...n..f...2....
000B60: 09 C1 10 E8 C9 95 F4 11 00 02 21 2F 17 CD F1 32 ..........!/...2
000B70: 14 1E 00 3E 12 BA 20 F2 C9 02 00 F0 DD 16 0F 05 ...>.. .........
000B80: 17 F0 23 17 0F 3C 17 0F 0A 17 0F 23 17 0F 3C 17 ..#..<.....#..<.
000B90: 0F 0F 17 F0 23 17 0F 3C 17 0F 14 17 0F 23 17 0F ....#..<.....#..
000BA0: 3C 17 0F AB 14 0F 19 17 F0 FB 14 0F FE 14 F0 0B <...............
000BB0: 15 0F 10 15 F0 23 15 0F 26 15 F0 33 15 0F 38 15 .....#..&..3..8.
000BC0: 00 00 41 44 44 52 45 53 53 20 4D 41 43 48 49 4E ..ADDRESS MACHIN
000BD0: 45 20 43 4F 44 45 20 20 20 20 53 4F 55 52 43 45 E CODE SOURCE
000BE0: 20 43 4F 44 45 20 20 20 20 A0 20 41 53 53 45 4D CODE . ASSEM
000BF0: 42 4C 45 44 20 43 4F 44 45 20 20 20 4C 41 42 45 BLED CODE LABE
000C00: 4C 53 20 20 49 4E 53 54 52 55 43 54 49 4F 4E 53 LS INSTRUCTIONS
000C10: 20 A0 43 75 72 73 6F 72 20 6B 65 79 73 2C 20 64 .Cursor keys, d
000C20: 65 6C 65 74 65 2C 20 61 6E 64 20 61 75 74 6F 2D elete, and auto-
000C30: 72 65 70 65 61 74 20 61 72 65 61 73 20 73 74 61 repeat areas sta
000C40: 6E 64 61 72 64 2E 20 45 53 43 20 72 65 74 75 72 ndard. ESC retur
000C50: 6E 73 20 79 6F 75 20 74 6F 20 6D 65 6E 75 2E 20 ns you to menu.
000C60: 20 A0 5B 43 DD 20 63 3D AC 6C 65 61 72 73 20 61 .[C. c=.lears a
000C70: 6C 6C 20 A0 45 4E 54 45 D2 20 74 61 62 73 20 74 ll .ENTE. tabs t
000C80: 6F 20 6E 65 78 74 20 6C 69 6E 65 A0 5B 41 DD 20 o next line.[A.
000C90: 61 73 73 65 6D 62 6C 65 73 20 20 A0 53 50 41 43 assembles .SPAC
000CA0: C5 20 74 61 62 73 20 69 6E 20 6C 61 62 65 6C 20 . tabs in label
000CB0: 61 72 65 E1 20 50 72 6F 67 72 61 6D 20 68 61 73 are. Program has
000CC0: 20 73 75 63 63 65 73 73 66 75 6C 6C 79 20 61 73 successfully as
000CD0: 73 65 6D 62 6C 65 64 2E 20 20 20 20 50 72 65 73 sembled. Pres
000CE0: 73 20 52 75 6E 2C 20 45 64 69 74 2C 20 6F 72 20 s Run, Edit, or
000CF0: 45 53 43 20 74 6F 20 65 78 69 74 2E 20 20 20 20 ESC to exit.
000D00: 20 20 20 A0 A0 20 50 72 6F 67 72 61 6D 20 63 6F .. Program co
000D10: 75 6E 74 65 72 20 68 61 73 20 6A 75 6D 70 65 64 unter has jumped
000D20: 20 74 6F 20 6E 6F 6E 2D 76 61 6C 69 64 61 64 64 to non-validadd
000D30: 72 65 73 73 2E 20 45 44 49 54 20 6F 72 20 72 65 ress. EDIT or re
000D40: 52 55 4E 20 70 72 6F 67 72 61 6D 2E 20 20 20 20 RUN program.
000D50: 20 20 20 20 A0 20 59 6F 75 20 61 72 65 20 74 72 . You are tr
000D60: 79 69 6E 67 20 74 6F 20 72 C7 30 75 6E 20 63 6F ying to r.0un co
000D70: 64 65 20 69 6E 20 61 6C 6C 6F 63 61 74 65 64 73 de in allocateds
000D80: 74 6F 72 61 67 65 20 61 72 65 61 2E 20 52 45 54 torage area. RET
000D90: 20 6D 69 73 73 69 6E 67 3F 20 50 6C 65 61 73 65 missing? Please
000DA0: 20 45 44 49 54 20 A0 20 50 72 65 73 73 20 5B 52 EDIT . Press [R
000DB0: 5D 20 74 6F 20 72 75 6E 20 65 78 61 6D 70 6C 65 ] to run example
000DC0: 2C 20 5B 45 5D 20 74 6F 20 20 65 64 69 74 2C 61 , [E] to edit,a
000DD0: 6E 64 20 45 53 43 20 74 6F 20 72 65 74 75 72 6E nd ESC to return
000DE0: 20 74 6F 20 6D 65 6E 75 2E 20 20 20 20 20 20 20 to menu.
000DF0: 20 20 20 20 20 20 A0 20 50 72 6F 67 72 61 6D 20 . Program
000E00: 73 74 6F 70 70 65 64 2E 20 59 6F 75 20 6D 61 79 stopped. You may
000E10: 20 72 65 2D 52 55 4E 2C 20 45 44 49 54 20 20 6F re-RUN, EDIT o
000E20: 72 20 70 72 65 73 73 20 45 53 43 20 74 6F 20 72 r press ESC to r
000E30: 65 74 75 72 6E 20 74 6F 20 6D 65 6E 75 2E 20 20 eturn to menu.
000E40: 20 20 20 20 20 20 20 A0 46 4C 41 47 53 20 20 20 .FLAGS
000E50: 20 52 65 67 69 73 74 65 72 73 20 61 6E 64 20 43 Registers and C
000E60: 6F 6E 74 65 6E 74 73 20 20 20 20 D6 C6 53 54 41 ontents ..STA
000E70: 43 CB 20 43 3A 20 A0 20 5A 3A 20 A0 50 56 3A 20 C. C: . Z: .PV:
000E80: A0 20 53 3A 20 A0 20 20 20 20 20 20 20 20 20 20 . S: .
000E90: 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20
000EA0: 20 20 20 20 20 20 20 20 20 20 20 20 20 A0 00 00 ...
000EB0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000EC0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000ED0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000EE0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000EF0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F00: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F10: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F20: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F30: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F40: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F50: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F60: 00 00 00 00 00 00 00 00 00 00 00 00 00 7A 6E 00 .............zn.
000F70: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F80: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000F90: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000FA0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000FB0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000FC0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000FD0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000FE0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000FF0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
001000: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
001010: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
001020: 00 00 00 00 00 DD 21 3F 1A CD D9 19 CD 15 1A CD ......!?........
001030: 06 BB 32 43 1A 3E 42 CD 1E BB 28 05 ED 7B 59 14 ..2C.>B...(..{Y.
001040: C9 CD 19 BD CD 2A 1A 3A 43 1A FE 0D 20 13 DD 36 .....*.:C... ..6
001050: 00 00 DD 34 01 DD 7E 01 FE 10 20 CD DD 35 01 18 ...4..~... ..5..
001060: C8 FE F2 01 FF 00 28 4E FE F3 01 01 00 28 47 4B ......(N.....(GK
001070: 19 FE F1 01 00 01 28 40 FE F0 01 00 FF 28 39 FE ......(@.....(9.
001080: 43 CA A2 19 FE 7F 28 6B FE 41 C8 FE 20 38 9F FE C.....(k.A.. 8..
001090: 3A 38 0A FE 61 38 97 FE 7B 30 93 D6 20 CD C1 19 :8..a8..{0.. ...
0010A0: 3A 43 1A FE 20 20 0E DD 7E 00 FE 06 30 07 DD 34 :C.. ..~...0..4
0010B0: 00 3E 20 18 E8 01 01 00 DD 36 03 0F DD 7E 00 81 .> ......6...~..
0010C0: FE 16 38 0C CB 79 20 05 3E 00 04 18 03 05 3E 15 ..8..y .>.....>.
0010D0: DD 77 00 DD 7E 01 80 FE 10 30 06 DD 77 01 C3 BA .w..~....0..w...
0010E0: 18 CB 78 28 07 DD 36 01 00 C3 BA 18 DD 36 01 0F ..x(..6......6..
0010F0: C3 BA 18 DD 35 00 F2 9A 19 DD 36 00 15 DD 35 01 ....5.....6...5.
001100: F2 9A 19 DD 36 00 00 DD 36 01 00 3E 20 CD C1 19 ....6...6..> ...
001110: C3 BA 18 CD B3 19 DD 36 00 00 DD 36 01 00 DD 36 .......6...6...6
001120: 02 00 18 94 21 45 17 11 46 17 36 20 01 6F 01 ED ....!E..F.6 .o..
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DDRESS MACHINE CODE SOURCE CODE ASSEMBLED CODE LABELS INSTRUCTIONS Cur
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may re-RUN, EDIT or press ESC to return to menu. FLAGS Registers and
Contents STAC C Z PV S zn!2CB(Y*C
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ionthe numbertoF)))*r&&&&&&'W*p****-++'C'+++5,,,p','''(2(c(,-.-b-(£(--.'.....6/F
////0F0v0(000121d11+)4))#######$$$($3$#1$T$$y$$$$$$$$$%%%E%S%a%g%%%%%%%&&&$&)&.&
3&8&&B&H&N&T&Z&&f&l&accumulatorthe numberHL registerlocation (HL)DE registerB re
gisterC registerif Carry flag iGs set,D registerE registerH registerL registerBC
registerif Zero flag is set,NOT DEFINEDif Zero flag is not set,if Carry flag is
not set,IX registerIY registerlocation (IX+offset)location (IY+offset)location
(BC)location (DE)AF registersalternative AF registersStack Pointerif Sign flag i
s 0, ie Positive,location (IX)location (IY)stackif Sign flag is 1, ie Minus,if P
/V flag is nOt set,if P/V flag is sEt,Interrupt vector registerRefresh registerp
ort whose number is in Cport with above numberFlag register0 of1 of2 of3 of4 of5
of6 of7 of0000H0008H0010H0018H0020H0028H0030H0038His loaded withhas added to it
has added to it carry andThe Accumulator has subtracted from ithas taken from it
carry andis INCremented by 1is DECremented by 1is PUSHed onto stack and SP decr
emented by 2is POPped from stack and SP incremented by 2Accumulator is logically
ORed withAccumulator is logically ANDed withBC, DE, HL registers are all EXchan
ged with the alternative setis EXchanqged withAccumulator is NEGated, ie made ne
gativecarry flag is inverted, ie Compliment Carry Flagcarry flag is set, ie Set
Carry Flagdecimal Adjust AccumulatorNo OPeration is performed, inspite of this i
t takes .000001 secs to do itcomputer is HALTed until an interrupt comes to awak
e it againtest BITSET bitRESet bitblock transfer (BC) bytes starting at (HL) to
(DE) incrementingtransfer (HL) to (DE), increment HL and DE, dec BC, reset PV fl
ag if BC0block transfer (BC) bytes starting at (HL) to (DE) decrementingtransfer
(HL) to (DE), decrement HL, DE, and BC, reset PV flag if BC0perform relative ju
mp todecrement B, if zero continue, if Not Zero perform relative jump topush the
return address to stack and CALLperform jump toRETurn from Non-maDRskable inter
rupt NOT PERFORMED BY SIMULATORRETurn from Interrupt NOT PERFORMED BY SIMULATORR
ETurn address is POPped from the stacksearch from (HL) Incrementing BC bytes for
data in Accumulatorcompare (HL) with A, increment HL, decrement BCsearch froom
(HL) Decrementing BC bytes for data in Accumulatorcompare (HL) with A, decrement
HL, and BCinvert all bits in A register, ie ComPLimentaccumulator is ComPared w
ithcontents of Accumulator is eXclusively ORed withrotate Accumulator Left shift
ing bit 7 into Carryrotate Accumulator and carry Leftrotate Accumulator Right, s
hifting bit 0 into Carryrotate accumulator and carry Rightrotate Left Decimal co
ntents of (HL) with low nibble of Accumulatorrotate Right Decimal contents of (H
L) with low nibble oLf Accumulatorchange Interrupt Mode toreSTart at addressiNpu
t from port (C), into block from (HL) Incrementing HL for B bytesiNput into (HL)
from port (C) increment HL, decrement BiNput from port (C), into block from (HL
) Decrementing HL for B bytesiN%put into (HL) from port (C) Decrement HL, decrem
ent Binput data fromoutput from (HL) to port (C) increment HL, decrement BouTput
from (HL) to port (C) block from (HL) Incrementing for B bytesoUTput from (HL)
to port (C) Decrement HL, decrement BouTput3 to port (C), from block from (HL) D
ecrementing HL for B bytesoUTputdisable Interrupts. NOT PERFORMED BY SIMULATOR.e
nable Interrupts. NOT PERFORMED BY SIMULATOR.is Rotated Left shifting bit 7 into
Carryand Carry are rotated Leftis Rotated Right, shifting bit 0 into Carryand C
arry are Rotated Rightis Shifted Left Arithmetically, 0 enters at bit 0is Shifte
d Right Arithmetically, sign bit unchangedis Shifted Right Logically, 0 is shift
ed into bit 7X!1v(.nf! S2 E0!EV(!hF"M2L22 E2,E2!Eo0$#fo( 2!123)E*( !1N(23 213(J
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abel too longNumber too largeLabel not foundNo DEFB, or DEFWOffset missing Error
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!B213## 2 DFE,E.EFLAGS Registers and Contents STAC You are about to
affect memory area notallocated to you. Please edit program. You are about to
write to memory which will affect your program. Please edit. The program has be
en completed. You mayEDIT or reRUN. Exit by pressing ESC. The stack pointer i
s outside allocated memory area. Please edit program. There are too many r
egisters called up,to display. Program will RUN as normal.qIIPkkkkPPAAkkkkkkkkkk
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RR0EX0LDLDILDLDDCPCPICPCPDDA'0CP/0NEDCC0SC70NO0RLoRRgININIININDOUT£OTIOUTOTDD0E0
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Q*XKXBQ* RP" R*X##"XP!X"X!" RP2( EN#(+0!6y7n7* RPo# 2A22 Press SPACE to
continue. Press SPACE to select & ENTER to start Please confirm loading r
equired Y/NW& W$ UX.X/X3X7XDXQXXeXoXvXXXXXXXXXXXXtXwNavbWcehk&loo ptsv1wzwzjM
(z £ REGISTERS and MEMORY Registers E are like pigeon holes inwhich
numbers can be stored. A singleregister can store any number between0 and
255. The Z80 has many suchregisters. Initially we shall consideronly t
he more general ones designatedA,B,C,D,E,H and L. The A register ( orAccumul
ator ) is the most important,since more instructions involve thisregiste
r than any other. The step by step instructions making upa program are stored in
memory, a seriesof similar registers or memorylocations external to
the Z80 chip.65536 memory locations can be accessed.They are designated
by their numberbetween 0 and 65535. It takes the Z80longer to access mem
ory than its owninternal registers.£ SIMPLE LOAD INSTRUCTIONS Perhaps a
better ' description of loadthan the conventional one is copy, sincea load ins
truction copies the numberheld in a register or memory locationinto anot
her register or memorylocation. As with all copying theoriginal re
mains unchanged. The Kload instruction is normallyabbreviated to LD. The
simplest form ofload copies data from one register toanother, i.e. LD A,B .
In these abbreviationed instructions( mnemonics ), it is conventional to put
the register affected first, followed bythe register from which the information
is copied. Hence LD A,B copies thecontents of B into A or Accumulator.£ We
can also load any of the registerswith a number, n, between 0 and 255dir
ectly , with instructions likeLD C,123. We shaRll discuss later fromwh
ere the number is copied. Suffice tosay at this stage, that instructionsar
e stored as numbers. i.e. LD A,B isstored as 78. Some instructions likeLD
A,B are stored as one number, otherstake up to four. The two types of instruc
tions discussed( LD r,r' and LD r,n ) can involve anyof the registers conside
red. Only theA register can be loaded with thecontents of a memory
location. Theinstructions of the form LD A,(nn) loadsor copies the content
s of memorylocation nn ( a number between 0 and65535 ) into the A regist
er.£ We can also load a memory location withthe contents of the Accumulator
withLD (nn),A where nn is the number oraddress of the memory location. All
other single regis ters cannot beloaded directly from a memory location.Two
instructions involving the Aregister are required.£ Summary - Loading
single registersLD r,r' where r and r' are any of the following A,B
,C,D,E,H and L.LD r,n where n is a number 0 - 255LD A,(nn) where nn is a nu
mber 0 to 65535LD (nn),A£ REGISTER PAIRS The fact that a
single register canonly hold numbers up to 255 and thetotal memory ava
ilable is up to 65535,is a limitation. For0x this reason thereare a whole s
et of instructions on theZ80 that deal with registers in pairs.The pairs are
BC, DE, and HL registerpairs. The two registers hold different partsof a numb
er. Consider the decimal number27. We can think of this as X having twoparts t
he high part ( or byte) is the 2since it represents 2*10 , the low part( or by
te ) is the 7, since it onlyreplesents 7*1 . The total number is2*10 + 7
*1 27. Each digit can only be0 - 9 i.e. 10 different numbers.£ A single reg
ister can hold 0-255, 256different numbers. If we put tworegisters side
by side, i.e. HL , the Hregister holding the High byte and the Lholding the L
ow byte, then we can holdin the register pair H*256 + L*1 in thesame way as 27
2*10 + 7*O1. The maximumthat can be held in a register pairis therefore
255 * 256 + 255 65535.By convention the high byte is storedin the first
register in the registerpair name. We can load any of the register pairswith
a number between 0 - 65535 directlyusing instructions like LD HL,nn. Just as we
can combine two registers tohold numbers up to 65535, we can combineadjacent
memory locations. Byconvention the Low byte is held in thelocation wit
h the Lower address.£ We can therefore load a register pairwith the contents
of a pair of memorylocations, with instructions likeLD DE,(nn) . Thi
s instruction isequivalent to the non-valid instructionsLD E,(nn) and LD D
,(nn+1). Similarly, we can load the contents ofa reg$ister pair into a pair o
f memorylocations, i.e. LD (nn),BC which isequivalent to the two
non-validinstructions LD (nn),C and LD (nn+1),B. There are no instructions
to loadregister pairs with the contents ofanother register pairk. Two in
structionsthe form LD r,r' are usually used toperform such an operation. The
re is one instruction similar tothis type EX DE,HL. It EXchanges theconte
nts of the DE register pair withthe contents of the HL register pair.£ Summa
ry - loading register pairsLD dd,nn where dd is any register pair BC
, DE, and HL. nn is a number 0 - 65535LD dd,(nn)LD (nn),ddEX DE,HL
exchanges register contents£ INDIRECT ADDRESSING Up to now we
have onFly usedinstructions in which memory locationshave been specified
directly in theinstruction. Another useful method ofspecifying a memory l
ocation is to use anumber held in a register pair, knownas indirect addressin
g. The instructionLD B,(HJL) for example allows us to loador copy to the B r
egister the contentsof the memory location whose address isin the HL register p
air. All single registers can be loadedusing the HL register pair as a poin
ter. Similarly , memory locations canE beloaded indirectly from any s
ingleregister using the contents of the HLregister pair as address i.e. LD (H
L),C.£ Use of the BC and DE registers forindirect addressing is limited to
the Aregister. i.e. LD (DE),A LD A,(BC) etc.£ Summaryb - indirect addressing
LD r,(HL) where r is any single register A,B,C,D,E,H, or L.LD (HL),rLD
A,(BC)LD A,(DE)LD (BC),ALD (DE),A£ ADDITIONS AND THE CARRY FLAG Both single
register and register pairaddition are possible on the Z80. All single reg
ister additions areperformed with the Accumulator. A number( i.e. ADD A,6 ),
the contents of aregister (i.e. ADD A,B), or the contentsof an indirectly
addressed memorylocation using the HL register pair(i.e. ADD A,(HL) )
can be added to theAccumulator. The result is held in theAccumulator. The s
ource of the additionis unaffected. Register pair addition is performed inconju
nction with the HL register pair,and can only involve BC or DE ( i.e.ADD H
L,BC or ADD HL,DE ). Again thethe result is held in the HL registerpair,
and the other register pair isunaffected.£ Single and double register ad
ditionswill obviously only give the correctresult if the answer is less t
han themaximum number the rWegister(s) can hold.If it is greater than this a c
arry isgenerated. The processor holds thissingle bit of information or
flag, sothat action can be taken over theoccurrence of the carry. If th
ere hasbeen a carry on the last arithmeticoperation the carry is said to b
e set toa "1" . If not set it is said to be a"0". A second form of addition i
s availableon the Z80 , for both single andmultiple registers, known
as add withcarry abbreviated to ADC. They aresimila r to ADD except that
if the Carryflag is set before the addition theresult is incremented by
one. Allpreviously mentioned ADD instructionscan be performed as ADC.£ AD
C instructions can be strung togetherto perform the addition of two numbersof
any length, as the examples willshow.£ Summary - additionADD A,n wher
e n is a number 0 - 255ADD A,r where r is any single registerA
DD A,(HL)ADD HL,BCADD HL,DEADC A,nADC A,rADC A,(HL)ADC HL,BCADC HL,DE£ SUBTRA
CTION AND THE CARRY FLAG Single register subtraction takes placewith the Accumul
ator. All the forms ofADD can be used in subtract. Theabbreviation or
mnemonic SUB is alwayswritten without the A, which is implied.Hence the instru
ctionas are SUB n, SUB r,and SUB (HL). Again the result is heldin the Accumula
tor. The carry flag isset to a "1" if the result is outsidethe range 0 -
255.There are no double register SUBinstructions.All configurations of
ADC instructionscan be used with SBC or SuBtract withCarry instructions. T
he operation issimilar to SUB, except that the resultis decremented by one
if the carry flagis set.£ Like ADC instructions SBC instructionscan be strung
together to subtract anylength number. Since double register subtraction cano
nly be performed with carry, the stateof the carry flag prior to SBC HL,BC,a
nd SBC HL,DE is important, and shouldbe "0". The carry flag can be set to a "1
" withthe instruction SCF or Set Carry Flag.Although there are no sp
ecificinstructions to clear the carry flag to"0" there is an instruction to inv
ert itCCF or Compliment Carry Flag. We shallsee later that all logic instructi
ons doclear the carry flag.£ SUMMARY - subtract with carrySUBv n )SUB r
) SUBtract from A, n, r, orSUB (HL) ) (HL)SBC A,n )SBC A,r
) SuBtract from A with carrySBC A,(HL))SBC HL,BC ) SuBtract from HL with ca
rrySBC HL,DE )SCF Set Carry FlagCCF Compliment Carry Flag£ INCREM
ENT AND DECREMENT INSTRUCTIONS The last arithmetic instructions to bediscussed
can be performed on any singleor double register(s). These are INC andDEC. INC
increments (or increases) thecontents of the register, or memorylocation i
ndirectly addressed by the HLregister pair, by one. DEC decrements( or decr
eases ) the contents of theregister or memory location indirectlyaddressed
by the HL register pair, by The carry flag is not affected byeither of
these instructions. These instructions are primarily usedwhere counters ar
e required. Theoperation on register pairs is alsouseful in sequenti
al operations onmemory locations utilising indirectaddressing.£ Summary
- increment and decrementINC rINC (HL)INC ddDEC rDEC (HL)DEC dd£ THE Z
ERO FLAG Another very useful flag included inall processors is the Zero fla
g. Thisflag is set to "1" if the result of anysingle register arithmetic opera
tion iszero. Otherwise it is cleared nto "0". It is only affected by double re
gisterarithmetic operations involving thecarry flag , i.e. only affect
ed byADC HL,dd or SBC HL,dd. The zero flag, like the carry flag, isunaffecte
d by any LD or EX instructions.£ Summary - zero andE carry flagsinstruction
Carry Zero r dd r dd LD . .
. . EX n/v . n/v . ADD * * * .
ADC * * * * SUB * n/v * n/v SBC
* * * * INC . . * . DEC .
. * .r single register dd double register * flag affected . fla
g not affected n/v instruction not valid£ Lesso'ns 1 - 9 1. R
egisters and Memory 2. Simple Load Instructions Ex - simple load instru
ctions 3. Register Pairs Ex - register pair loading 4. Indirect Addres
sing Ex - indirect addressing 5. Addition and the Carry Flag Ex
- single register addition Ex - register pair addition 6. Subtraction
and the Carry Flag Ex - register subtraction Ex - register pair
subtraction 7. Increment and Decrement Ex - inc and dec instructions 8.
, Zero Flag Ex - zero flag 9. Compare Ex - comparison instructio
ns Load further lessons from tape£ INTRODUCTION The Z80 is the micro
processor at theheart of your Amstrad. This program willteach you the use
of all £ theinstructions available on the Z80, ina series of simple pr
ogressive lessons. After each lesson, step by stepexamples are given. T
o further improveyour understanding of the topic, you maythen modify or rewrite
the examples, andrun them pwithout fear of crashing thesystem. Proceed thro
ugh the menu by pressingENTER to begin the highlighted lesson orexample, and
SPACE to jump to the next.At anytime BREAK will return you to themenu.£ INTR
ODUCTION TO RUNNING SIMULATOR All examprjles are run using theSimulator
. At this stage the only areasof the simulator screen to consider arethe Sourc
e Code ( the third column thatholds the instructions ) and the area atthe botto
m of the screen, which displaysthe contents of the regisv%ters. Otherareas
will be described as they arerequired. The assembler, that converts instruc
t-ions into machine code, will acceptLabels. Instead of putting an addre
ssinto the program, we can give it a name.The assembler will then allocate
amemory location to it. The name must becalled up in the program with i
tsinitial condition using the instructionDEFB, DEFine Byte.£ DEFB is an assemb
ler instruction not aZ80 instruction. The allocated memoryaddress is given in
the first column. The simulator will display the contentsof this memory locati
on, in decimal, inthe second column. Before each instruction is executed theinst
ruction is described in ENGLISH. Tryto work out what should happen beforepres
sing any key to perform theinstruction.£ You may go through the exampl
e as manytimes as you like, until you understandit. If you still find difficul
ty, pressESC, skip over the lessons and re-ENTERthe lesson with which you ar
e havingdifficulty. The manual however gives anoutline of each lesson. When y
ou do understand the examplesgiven try modifying them using theeditor.
Then try entering your ownprograms. All programs should end withRET ( or
return ). Don't worry if youforget, the simulator will tell you. Remember,
it is impossible for aprogram to destroy your computer. You'llneed a hamme
r to do that!£ LD A,34 LD B,A LD (3867),A LD A,(STORE) LD (3867),A LD A,B
LD (STORE),A LD C,B LD A,(3867) LD A,67 LD (3866),A RET STORE DEFB 124 D
EFB 0 REGISTER PAIR EXAMPLES The example illustrates each type ofi
nstruction that loads register pairs.It also demonstrates that a register/m
emory pair are still two individualregisters or memory locatiV0ons and can b
etreated as such. Another assembler instruction isintroduced, the DEFW
instruction. Theassembler allocates 2 memory locationsto the associated lab
el. The numberafter the DEFW sets the initialconditions. The address
in the first column is thelow byte address. The second columndisplays th
e contents of the pair ofmemory locations as a single number.£ LD DE,256 LD
E,4 LD (STORE),DE LD (LOW),DE LD A,2 LD (HIGH),A LD HL,(LOW) EX DE,HL LD
D,0 RWET STORE DEFW 32000 LOW DEFB 0 HIGH DEFB 0 EXAMPLES OF INDI
RECT ADDRESSING These examples illustrate indirectaddressing. At this
stage it isimpossible to demonstrate its useful-ness. It is employed ext
ensive in laterexamples£ LD HL,LOW LD C,(HL) LD HL,HIGH LD B,(HL) LD A,(BC)
LD DE,3860 LD (DE),A LD (HL),0 LD BC,3862 LD (BC),A RET LOW DEFB 21 HI
GH DEFB 15 DEFB 0 EXAMPLES OF SINGLE REGISTER ADDITION The example adds
together Wg the twonumbers 8740 and 1260 using singleregister additi
on only. The doublelength numbers are held in BC and DE sothat the result
s can be displayedeasily. This method of addition can beperformed using r
egister pair addition,as wellD, to add together numbers of anylength. The stat
e of the individual flags aredisplayed to the left of the registers.£ LD DE,87
40 LD BC,1260 LD A,E ADD A,C LD C,A LD A,D ADC A,B LD B,A LD HL,STORE
LD A,15 ADD A,(HL) RET STORE DEFB 25 EXAMPLE OF REGISTER PAIR ADDITION
This example effectively multiplies anumber held in the HL register pair by
10 using addition to successivelymultiply by 2.£ LD HL,(STORE) ADD HL,HL
ADD HL,HL LD DE,(STORE) ADD HL,DE h ADD HL,HL LD (STORE),HL LD HL,1000
LD A,255 ADD A,1 ADC HL,HL RET STORE DEFW 6000 EXAMPLE OF SINGLE REGISTE
R SUBTRACTION This example subtracts 1260 from 8740using only single register
subtraction.The two numbers are held in allocatedmemory locations this time.
This method of subtraction can beextended to any length number and
register pair subtraction.£ LD HL,NUM1 LD DE,NUM2 LD A,(DE) SUB (HL) LD (DE)
,A LD E,18 LD L,20 LD A,(DE) SBC A,(HL) LD (DE),A RET NUM2 DEFW 8740 NU
M1 DEFW 1260 EXAMPLE USING REGISTER PAIR SUBTRACTION This example illustra
tes subtraction ofregister pairs. It subtracts 1536 from65536. At least 3 b
ytes of memory arerequired to hold the latter number,the highest byte repr
esenting 65536. Although ADC and SBC may require theCarry flag to be cleare
d first, if theprevious calculation NEVER results in acarry being generated, t
his step may beomitted. However it is often better tobe safe and include it,
than look forthe randomly occurring fault that it cangenerate.£ LD HL,(NUM1L)
LD DE,(NUM2L) SBC HL,DE LD (ANSL),HL LD HL,(NUM1H) LD DE,0 SBC HL,DE LD (
ANSH),HL RET NUM1L DEFW 0 NUM1H DEFW 1 *65536 NUM2L DEFW 1536 ANSL DEFW 0 AN
SH DEFW 0 EXAMPLE OF INCREMENT AND DECREMENT At present the programs th
at can bewritten are somewhat limited. INC and DEC really only become usefuli
n conjunction with instructions to beintroduced shortly.£ LD C,5 LD HL,STORE
LD (HL),C INC C INC HL LD (HL),C DEC C INC L LD (HL),C INC H RET STOR
E DEFB 0 DEFB 0 DEFB 0 EXAMPLES OF ZERO FLAG While running this
example try topredict the state of the Zero flag afterthe instruction has been
performed.£ LD HL,257 DEC L LD A,23 SUB 23 LD DE,256 INC A SBC HL,DE INC
H SUB 2 SBC HL,DE LD HL,1 DEC HL RET COMPARE So far
all instructions we havediscussed that affect the Carry orZero fPla
gs also affect the registerconcerned. There are a number of usefulinstruct
ions that only affect flags. Thecompare or CP instruction is one ofthese. C
P compares the contents of the Aregister with a number (CP n), anotherre
gister (CP r), or any memory locationindirectly addressed through the HLre
gister pair ( CP (HL) ). The compareinstruction is effectively SUB but there
sult does not affect the contents ofthe A register.£ If n is the number w
ith which A iscompared, then the following results - Carry
Zero A n 0 0 A n 0 1 A n
1 0£ Summary - comparisonsCP n Compares A with n ( 0
- 255)CP r Compares A with register rCP (HL) Compares A with memory locati
on (HL)£ EXAMPLES OF COMPARE Try to
predict the state of the Zeroand Carry flags before performing thecomparis
on. In the next lessons we willbe using tW5hese flags extensively, toproduc
e more interesting examples.£ LD A,5 CP 4 CP 5 CP 6 LD B,3 CP B LD HL,STOR
E CP (HL) DEC (HL) CP (HL) ADD A,230 CP (HL) RET STORE DEFB 6 0pp088((
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