The Complete Machine Code Tutor (UK) (Face 3) (1985) [Original] [TAPE] [UTILITAIRE].cdt
HexaDumpAscii only
000000: 5A 58 54 61 70 65 21 1A 01 0C 20 0D 40 11 09 09 ZXTape!... .@...
000010: A6 04 57 04 8F 04 1C 09 00 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 06 09 A6 04 57 04 8D 04 ..y.........W...
000130: 18 09 00 10 08 E7 21 11 06 00 16 CD A1 BC D4 00 ......!.........
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 08 09 A6 04 .....P..........
000750: 57 04 8B 04 16 09 FF 0F 08 FB 1D 8B 43 00 6C 0D W...........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 B2 s been stopped£.
000860: 00 20 20 20 20 20 20 20 50 6C 65 61 73 65 20 73 . 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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MACHINE CODE SOURCE CODE ASSEMBLED CODE LABELS INSTRUCTIONS Cursor ke
ys, delete, and auto-repeat areas standard. ESC returns you to menu. C clears a
ll ENTE tabs to next lineA assembles SPAC tabs in label are Program has success
fully assembled. Press Run, Edit, or ESC to exit. Program counter has
jumped to non-validaddress. EDIT or reRUN program. You are trying to r0u
n code in allocatedstorage area. RET missing Please EDIT Press R to run example
, E to edit,and ESC to return to menu. Program stopped. You may re
-RUN, EDIT or press ESC to return to menu. FLAGS Registers and Conte
nts STAC C Z PV S zn!2CB(Y*C 64 5(N
(GK((9C(kA 88 a80 C 04 68y w0wx(6656566 666!EF6 oO!E(Gq!C!E2##!Cn,,go#og#g!1V(o
&)))%)r( (eoOw#wwwwwww999999w3w"3q33ww3ww3w333w33wwwwwwww3w3w333"33wwwwww33333ww
wfww33333x(33www33333wwwwwwww3ww3w3w33ww33"""33w333333333w33wwwwwww33333wwwww333
3333w3w33"333"333w33333w3ww333"U33333333Uwwww33 3w333"333ww3333ww33w3w333333333w
33333D333333333333333ww33wwwww333w3wwwwww3U3""3"w33"33333333"w3333333w3""333w333
3w2!1(0V (!" "!!80( ! !!(T(((j( !!! !!" "!! !!!""0!!#"!!#"!!/#"!(v(r(t(9(!!/#
"(((" !# " "!" ""8 E nf "!" "g" E(8(t(r(H(J(2(!" "(v(!/#"!" "!" "!" "(".E!z8 E
,ON#fi1o(21 E#((E # (+ # E &8!3$8!13addressthe contents ofmemory locationthe
numbertoF)))*r&&&&&&'W*p****-++'C'+++5,,,p','''(2(c(,-.-b-(£(--.'.....6/F////0F
0v0(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 register
C registerif Carry flag iGs set,D registerE registerH registerL registerBC regis
terif Zero flag is set,NOT DEFINEDif Zero flag is not set,if Carry flag is not s
et,IX registerIY registerlocation (IX+offset)location (IY+offset)location (BC)lo
cation (DE)AF registersalternative AF registersStack Pointerif Sign flag is 0, i
e Positive,location (IX)location (IY)stackif Sign flag is 1, ie Minus,if P/V fla
g is nOt set,if P/V flag is sEt,Interrupt vector registerRefresh registerport wh
ose number is in Cport with above numberFlag register0 of1 of2 of3 of4 of5 of6 o
f7 of0000H0008H0010H0018H0020H0028H0030H0038His loaded withhas added to ithas ad
ded 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 decremente
d by 2is POPped from stack and SP incremented by 2Accumulator is logically ORed
withAccumulator is logically ANDed withBC, DE, HL registers are all EXchanged wi
th the alternative setis EXchanqged withAccumulator is NEGated, ie made negative
carry flag is inverted, ie Compliment Carry Flagcarry flag is set, ie Set Carry
Flagdecimal Adjust AccumulatorNo OPeration is performed, inspite of this it take
s .000001 secs to do itcomputer is HALTed until an interrupt comes to awake it a
gaintest BITSET bitRESet bitblock transfer (BC) bytes starting at (HL) to (DE) i
ncrementingtransfer (HL) to (DE), increment HL and DE, dec BC, reset PV flag 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 jump tod
ecrement B, if zero continue, if Not Zero perform relative jump topush the retur
n address to stack and CALLperform jump toRETurn from Non-maDRskable interrupt N
OT PERFORMED BY SIMULATORRETurn from Interrupt NOT PERFORMED BY SIMULATORRETurn
address is POPped from the stacksearch from (HL) Incrementing BC bytes for data
in Accumulatorcompare (HL) with A, increment HL, decrement BCsearch froom (HL) D
ecrementing BC bytes for data in Accumulatorcompare (HL) with A, decrement HL, a
nd BCinvert all bits in A register, ie ComPLimentaccumulator is ComPared withcon
tents of Accumulator is eXclusively ORed withrotate Accumulator Left shifting bi
t 7 into Carryrotate Accumulator and carry Leftrotate Accumulator Right, shiftin
g bit 0 into Carryrotate accumulator and carry Rightrotate Left Decimal contents
of (HL) with low nibble of Accumulatorrotate Right Decimal contents of (HL) wit
h low nibble oLf Accumulatorchange Interrupt Mode toreSTart at addressiNput 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) Decr
ementing HL for B bytesiN%put into (HL) from port (C) Decrement HL, decrement Bi
nput 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 por
t (C) Decrement HL, decrement BouTput3 to port (C), from block from (HL) Decreme
nting HL for B bytesoUTputdisable Interrupts. NOT PERFORMED BY SIMULATOR.enable
Interrupts. NOT PERFORMED BY SIMULATOR.is Rotated Left shifting bit 7 into Carry
and Carry are rotated Leftis Rotated Right, shifting bit 0 into Carryand Carry a
re Rotated Rightis Shifted Left Arithmetically, 0 enters at bit 0is Shifted Righ
t Arithmetically, sign bit unchangedis Shifted Right Logically, 0 is shifted int
o bit 7X!1v(.nf! S2 E0!EV(!hF"M2L22 E2,E2!Eo0$#fo( 2!123)E*( !1N(23 213(J (E2
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2 (,(y((B*(8 9y8(&($( &6666 7h!- +w!!( !(S12!2K(()(! (E,(Aa708&0"Q Q(s (,(!7A8G8
H Sx!N7y x!a A 787e8Ah(PJP( 8r1((8)(/t12186) v2y##"!7*ut (,()(w#!m71 w#21 w#Y(
8o&!8(NF8 ##"!7!6Y(a(8g(!8g(#68g(8g(gg0 -9£ (7# ( y2!Eo0$#fo*K1+#q#p#"!4w3#! !6!
6n!7nfG(!7# ( (#VGut srnf##RuSpace missingInstr. ,1unknownMissing space or ,Erro
r after inst.Offset too bigCan only ADD IX/INumber missingBracket missingLabel t
oo longNumber too largeLabel not foundNo DEFB, or DEFWOffset missing Error foun
d during assembly. Pleasecorrect before running program. d's EaQ2)c)!
B6!CB! (B(##6#6#6#6# 2!5D2!"B!B6#N#Fx#8 0xG#w#r#6#6x!EN#fi22))!CB(V(133!BF %v
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+w"Bnf"1*AQAA*BZB(*BN#F#"BC1*A!C27nfk!CBnfknfAk!CBnFfGk*Bk*Bk!CBnFfGNFAk!CBnFfGF
ANBk!CBnfN k!CBnfN Bk! ,8',0!CB(r0!EC!B272 C!CBfnwutwfxnvuxtv!CB!BN qw #!6#"*
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# 2 DFE,E.EFLAGS Registers and Contents STAC You are about to affec
t 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 been com
pleted. You mayEDIT or reRUN. Exit by pressing ESC. The stack pointer is outs
ide allocated memory area. Please edit program. There are too many registe
rs called up,to display. Program will RUN as normal.qIIPkkkkPPAAkkkkkkkkkkkF!kkk
kkkkkkPPAAAAAAAAAAAAAkAPPPPPPPkyEEEEEzEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
EEEEEEEEHDBNNNIIA(SPSPP(C(HL(NN(NDI(IX(IY(BC(DEAF(IX(IY0008101820283038OOOOFHHvH
9HIHFIIIIPPIIJOOOOPJBJlJJOJPP PvOOOOO*P4PPOPORPHPfPPNpPzPPPOPPPPPPJJKDKjKKKKLML0
0200w060 0000&0.0!000*0K"0SC F0 p0N0q0V0r00s0 x0y0z0000 G0O0W00g0o0, 0.041044s,0
.040 A0 B0 C0 D0 E0 H0J0K0L0M0 P0Q0S0T0U0 X0Y0Z000 0a0b0c0e0 h0i0j0k0l0n0f0t0u0D
WDGFFO0 0I0R00d0m0kcAD0 0000000000)0490AD00JZ 0000004zj0SU00 0000000SB00BR 0000
004rb0DE050+0 000-0%07Z00 040IN040#0 0000$0,000430J080"00(0 0J0 00"0006000B0DJN
0CAL0"0 0006000B0RE00 00"06000B0PUS0000!0PO00000C00 0000000O00 0000000XO00 00000
00AN00 000£0000E00200RL 00000000R nt00000000RR 00 000 000R 00000000SL'0 0!0"0#0
&0%0$0SR/0 (0)0*0+0.0-0,0SR0 809000000BING0PO0RW0T0Vg0Xo0Zw0/0NF0PN0RV0T0Vf0Xn0Z
v00N 0P H0R P0T X0V 0X h0Z p0 x0NA0PI0RQ0TY0Va0Xi0Zq0y0NB0PJ0RR0TZ0Vb0Xj0Zr0z0NC
0PK0RS0T0Vc0Xk0Zs00ND0PL0RT0T0Vd0Xl0Zt00NE0PM0RU0T0Ve0Xm0Zu00SEN0P0R0T0V0X0Z00N0
P0R0T0V0X0Z00N 0P 0R 0T 0V 0X 0Z 0 0N0P0R0T0V0X0Z00N0P0R0T0V0X0Z00N0P0R0T0V0X0Z0
0N0P0R0T0V0X0Z00N0P0R0T0V0X0Z00REN0P0R0T0V0X0Z00N0P0R0T0V0X0Z00N 0P 0R 0T 0V 0X
0Z 0 0N0P0R0T0V0X0Z00N0P0R0T0V0X0Z00N0P0R0T0V£0X0Z00N0P0R0T0V0X0Z00N0P0R0T0V0X0Z
00IJ0Hx H%HPHXHHhHHLHpOUJ0HyH AHQHYHaHiHIINFPVRRS00b0d0f0h0j0l0RLC0RL0RRC0RR0EX0
LDLDILDLDDCPCPICPCPDDA'0CP/0NEDCC0SC70NO0RLoRRgININIININDOUT£OTIOUTOTDD0E0RETMRE
TEHALv0DEF0BI0DEF0Q!P" R3*XQ!7RQQ Q *XKXB !RQQS*XnfF nfN Q!RQB $ V(21"AQQ*XKXB
Q* RP" R*X##"XP!X"X!" RP2( EN#(+0!6y7n7* RPo# 2A22 Press SPACE to contin
ue. Press SPACE to select & ENTER to start Please confirm loading require
d Y/NWWUX&X'X+X2X9XXDXKXRXbXiXyXXXXXXbYXUa/cddevfmgijlmjnopsOurwxyzI! EX
AMPLES OF BIT MANIPULATION The following illustrates the way inwhich SET, RE
S, and BIT can be used tomanipulate individual bits of a registeror memory loca
tion.£ BIT MANIPULATION We have so far treated information inthe fo
rm of bytes or words. The Z80 doesallow a number of operations on separateor gro
ups of bits within one byte Any bit in a register, or memorylocation usin
g (HL), can be SET to a"1". We can also RESet any bit to a "0".To test the
state of a bit in aregister, or memory location, we can usethe instructi
on BIT. In this instructionthe Zero flag is set if the appropriatebit is "0"
and reset to a "0" if theappropriate bit is "1". Hence we canproduce our o
wn flags to indicate eventsand then act upon them later in ourprograms.£ Su
mmary - bit manipulationSET N,ir where N is the bit number 0-7SET N,(HL)RES N
,rRES N,(HL)BIT N,rBIT N,(HL)£ LD HL,STORE SET 6,(HL) LOOP INC (HL) BIT 2,(HL
) JR Z,LOOP RES 3,(HL) LD B,5 SET 7,B RES 2,B DEC B BIT 5,B RET Z STOR
E BIN 16 LOGICAL INSTRUCTIONS There are three logical instructio
nsavailable on the Z80. They are allperformed on a bit by bit basis betwe
enthe Accumulator and a number, register,or indirectly addressed memory locati
on(HL). The result is left in theAccumulator. In the AND instruction, i
f a bit in theAccumulator AND other number are "1",then the corresponding bit
in the resultwill be "1". If not it will be "0" i.e. 01101100 01
010110 ANDed together gives 01000100 As w'ell as performing ANDs on our
ownflags, this instruction is useful inmasking off areas of words, or resett
inggroups of bits in the Accumulator.£ In the OR instruction, if a bit in theAc
cumulator OR in the other number is a"1", the corresponding bit in the resultwi
ll be "1". If not it will be "0" i.e. 01101100 01010110 ORed to
gether give 01111110 As well as performing OR operations onour own flags
, this instruction isuseful in SETting a group of bits in theAccumulatHor.£
In the XOR (eXclusive OR) instruction,if the bit in the Accumulator is thes
ame as that in the number , then thecorresponding bit in the result will be"
0". If they are different it will be"1". Another way of looking at the XORi
nstruction is if one OR the other is"1" but NOT both, the answer will be"
1". i.e. 01101100 01010110 XORed together give 00111010 A
part from performing the logical XORfunction on our own flags the XORin
struction is useful in invertingindividual or groups of bits.£ The three l
ogical instructions AND, OR,and XOR affect the Zero, Sign, andParity flags
. The Carry flag is reset to"0" in all cases. Thus the limitation ofno Clear
Carry instruction can beperformedl by a logical instruction. AND A or OR A w
ill clear the carry flagand not affect any registers. XOR A willclear Carry and
the Accumulator.£ Summary - logical instructions AND n AND r AND (HL) OR n OR r
OR (HL) XOR n XOR r XOR (HL)£ EXAMPLEoS OF LOGICAL INSTRUCTIONS The examples
illustrate the operationof the logic instructions. The pupilshould test
his/her understandingfurther by modifying the contents of the3 stores.£ LD
HL,ST2 LD A,(ST1) AND (HL) LD (ST3),A LD A,(ST1) OR (HL) LD (ST3),A LD A,
(ST1) XOR (HL) LD (ST3),A RET ST1 BIN 65H ST2 BIN FH ST3 BIN 0H
SHIFT INSTRUCTIONS A shift instruction is one in which thebits of a re
gister or memory locationare moved sideways, left or right to theadjacent bit.
As we shall see, thisgives us a means of division as well asmultiplication
.£ The SRA r, and SRA (HL) instructionsshift the register/memory location (H
L)right, shifting bit 0 into the Carryflag and retaining the statYe of bit 7
. i.e. 7 6 5 4 3 2 1 0 C 0 1 1 0 0 0 1 0 x (98 dec.) becomes 0
0 1 1 0 0 0 1 0 (49 dec.) or 1 1 1 0 0 0 1 0 x (-30 dec.)becomes 1
1 1 1 0 0 0 1 0 (-15 dec.) Hence the SRA or Shift RightArithmetica
l8ly instruction divides bothpositive and negative numbers by 2.£ The SRL r,
and SRL (HL) instructionsshift the register/memory locationcontents right
, shifting bit 0 into theCarry flag and a "0" into bit 7. i.e. 7 6 5 4 3 2 1
0 C F51 0 1 1 0 1 1 1 x (183 dec.)becomes 0 1 0 1 1 0 1 1 1 (9
1 dec.) and Carry The SRL or Shift Right L
ogicallyinstruction therefore divides a positivenumber by 2. As with SRA the Ca
rry flagindicates the half.£ Theq SLA r, and SLA (HL) instructionsshift the
register/memory contents leftshifting a "0" into bit 0 and bit 7 intothe Carry f
lag. i.e. C 7 6 5 4 3 2 1 0 x 0 1 1 0 0 0 1 0 (98 dec.)becomes
0 1 1 0 0 0 1 0 0 (196 dec.) Hence the result of the instruction SLAor Shif
t Left Arithmetically is tomultiply the positive number by 2. TheCarry in
dicates a result greater than255.All these instructions affect not onlythe C
arry flag, but the Zero, Sign, andParity flags.£ Summary - shift instructions S
RA r divides +ve and -ve numbers by 2 SRA (HL) SRL r divides +ve numbers 0 - 2
55 by 2 SRL (HL) SLA r multiplies +ve and -ve numbers
by 2 SLA (HL)£ EXAMPLES OF SHIFT INSTRUCTIONS The6 three shift
instructions areillustrated using a fixed initial valuein the memory locatio
n STORE.£ LD HL,STORE LD B,5 LOOP1 SRA (HL) DJNZ LOOP1 LD (HL),A5H LD B,5 LO
OP2 SLA (HL) DJNZ LOOP2 LD B,8 LOOP3 SRL (HL) DJNZ LOOP3 RET £M STORE BIN 6
9H ROTATE INSTRUCTIONS These instructions are shift instruct-ion
s in which the bit that falls out oneend of the register or memory locationis
pushed back into the other end. TheCarry flag either forms part of thenum
ber shifted, thereby making it 9 bitslong, or duplicates the state of the bittha
t fell out of the register. There aretherefore four different types ofrot
ate.£ RLC r, and RLC (HL) (namely Rotate LeftCarry duplicating ), shifts the con
tentsleft, bit U7 rotating into bit 0, and theCarry duplicating the transferred
bit. i.e. C 7 6 5 4 3 2 1 0 x 1 0 1 1 0 0 0 1becomes 1 0 1 1 0 0
0 1 1 old bit 71 RL r, and RL (HL) Rotate Left, rotatesthe register/memory lo
cation with theCarry as ia 9th bit, left. i.e. C 7 6 5 4 3 2 1 0 c
1 0 1 1 0 0 0 1becomes 1 0 1 1 0 0 0 1 c cold Carry RL instructions can be
strung togetherto multiply any length number by 2,since the Carry is tra
nsferred betweenrepeated RL instructions.£ RRC r, RRC (HL) instructions (Rota
teRight Carry being duplicated) is similarto the RLC instruction but is a rig
htshift. Bit 0 is shifted into bit 7 andthe Carry duplicates the old bit 0. i.
e. 7 6 5 4 3 2 1 0 C 1 0 1 1 0 0 0 1 xbecomes 1 1 0 1 1 0 0 0 1
old bit 01 RR r, and RR (HL) instructions ( RotateRight ) is similar to RL excep
t that theshift around the 9 bits is to the right. i.e. 7 6 5 4 3 2 1 0 C
1 0 1 1 0 0 0 1 cbecomes c 1 0 1 1 0 0 0 1 cold Cfjarry RR instruction
s can be strung togetherto divide any length number by 2.£ RLC and RRC instruc
tions are useful insequentially interrogating the wholecontents of a registe
r without corrupt-ing its contents. All the above rotate instructions RLC,RL,
RRC, and RR not only affect theCarry flag but also the Zero, Sign andPari
ty flags, as have all shiftinstructions discussed. There are 4 other
rotate instructionsthat involve the Accumulator only. Theyare RLCA, RLA, RRCA
, and RR#gA. They areidentical to RLC A, RL A, RRC A, andRR A, but only aff
ect the Carry flag,and are twice as fast.£ Summary - rotate instructions RLC r
rotate r left, carry duplicates RLC (HL) RLCA rotate A left, carry duplica
tes RL r rotate r and carry left RL (HL) RLA rotate A and carry left RRC r
rotate r right, carry duplicates RRC (HL) RRCA rotate A right, carry duplicat
es RR r rotate r and Carry right RR (HL) RRA rotate A and Carry right£
EXAMPLES OF ROTATE INSTRUCTIONS These examples illustrate the operationof rota
te instructions. Modify theinitial condition of the location STOREto inves
tigate their operation further.£ LD HL,STORE LD B,8 LOOP1 RLC (HL) DJNZ LOOP1
LD B,8 LOOP2 RL (HL) DJNZ LOOP2 LD B,8 LOOP3 RRC (HL) DJNZ LOOP3 LD B,8 LOO
P4 RR (HL) DJNZ LOOP4 RET STORE BIN 69H EXAMPLE USING ROTATE AND SHIFT T
his example uses shifts and rotateinstructions to multiply two 8 bitnu
mbers together. One number is rotatedto examine each bit in turn. The othernu
mber is also shifted and added to thetotal if the bit is set. This method isfa
r quicker than the method of repeatedaddition.£ LD HL,0 LD DE,(NUM2) LD A,(NU
M1) LOOP RR A JR NC,JP1 ADD HL,DE JP1 RET Z SLA E RL D JR LOOP NUM1 D
EFB 212 NUM2 DEFB 203 DEFB 0 DECIMAL ROTATE The Z80 allows
us to rotate left andright nibbles as well as bits. RLD Rotates the Decimal n
umber Left RRD Rotates the Decimal number Right Th e best way to descri
be theseoperations is by example. Bothinstructions involve the Acc
umulator,and the memory location (HL). Theexamples are in Binary Coded D
ecimal ofcourse, each digit taking up one nibbleor four bits of tnhe register
or memorylocation.£ A after (HL) A before
5 4 x 3 after RLD x 5 4 3where x is any number and is unaffec
tedby the instruction. As can be seen the effect of stringingtogether RLD instr
uctions is to multiplya decimal number by 10. It is alsouseful in isolatin
g the high digit fromthe two digit number.£ A after (HL) A bef
ore x 5 4 3 after RRD 5 4 x 3 The RRD ins
truction therefore dividesthe decimal number by 10. It can also bestrung togethe
r to divide any length BCDnumber by 10. The low digit of the twodigit decimal
number can be isolated bythis instruction.£ Summary - decimal shiftsRLD Rota
te Left Decimal (*10)RRD Rotate Right Decimal (/10)£ EXAMPLES OF DECIMA
L ROTATION These examples illustrate themultiplication and divisio
n of BCDnumbers by ten.£ LD HL,STORE LD A,0 RLD INC HL RLD LD A,0 RRD
DEC HL RRD RET STsORE DEFW 254H INTRODUCTION This group
of lessons will introducethe idea of bit manipulation and its usein flag op
erations and in producingfaster methods of multiplication. But first we shal
l look further at theprocessors fl4ags.£ LESSONS 18 - 25 18. Parit
y 19. The flag register and AF 20. S and P/V flags in instructions
Ex - sign and parity flags 21. Bit manipulation Ex - bit manipulati
on 22. Logical instructions Ex - logical instructions 23. Shift inst
ructions Ex - shift instructions 24. Rotate instructions Ex
- rotate instructions 1 Ex - rotate instructions 2 25. Decimal rotat
ing Ex - decimal rotation Load further lessons from tape£
PARITY When data is transmitted from one placeto another, it is possible for
errors toappear in the received data. Parity is asimple way of detecting single
errors indata. An additional bit is added to thedata that makesI the total num
ber of "1"sin the word an even number. If thereceived word does not ha
ve an evennumber of "1"s in it, then an error hasoccurred. In addition to
this evenparity we could use odd parity, wherethe word is made to have an
odd numberof "1"s in it. The Z80 has a flag that is set if thenumber of "1"s
in a number is even. Thisflag is in fact the same flag as theoverflow flag.
It is termed the P/Vflag.£ i.e.0 0 1 1 0 1 0 0 parity flag is not set
s'ince number of "1"s is 3. i.e. odd.0 1 0 0
0 1 1 1 parity flag is set since number of "1"s
is 4. i.e. even. The instruction performed determinesthe meaning
of the P/V flag. Of all w$the instructions described todate only DAA treat
s it as a parityflag. All others either have no effecton it or use it as a
n overflow flag.All subsequent instructions either haveno effect on it or use
it as parity.£ THE FLAG REGISTER AND AF REGISTER PAIR We have discussed flags
as individualbits. This is the way they are normallyused. They are however a
ctually storedin the Z80 in a register, designated asF or flag register. Th
e flags arearranged as follows - bit 0 C JCarry bit 1 N
Subtract last bit 2 P/V Parity and overflow bit 3 Not use
d bit 4 H Half carry bit 5 Not used bit 6 Z Z
ero bit 7 S Sign£ The Flag register sometimes forms 1 aregister
pair in conjunction with theAccumulator. This pair is called AF. Theonly instr
uctions previously mentionedthat involve this register pair are PUSHand POP.£
S and P/V FLAGS IN INSTRUCTIONS The sign and parity flags can be usedin a
ljl absolute ( NOT relative) jumps,calls, and returns. The notation is - JP PO,
nn CALL PO,nn RET PO Parity odd0 JP PE,nn CALL PE,
nn RET PE Parity even1 JP P,nn CALL P,nn RET P
Sign +kCve JP M,nn CALL M,nn RET M Sign -ve If the condition is not met th
e programwill not jump, call a subroutine, orreturn.£ EXAMPLE OF JUMP
ON SIGN This example is a simple program toillustrate JP P,nn , and JP M,n
n jumpif positiveh and jump if minus. Note thatthe carry is still operativ
e, andtherefore, if you do not clear it beforeADC and SBC the later jump c
ould beunexpected. In this example the carry isnot cleared, so "look before you
leap".£ LD HL,52 LD DE,33 LD B!oC,5 LOOP1 SBC HL,DE LD (LREG),HL JP P,LOOP1
LOOP2 ADC HL,BC LD (LREG),HL JP M,LOOP2 RET Z JR LOOP1 LREG BIN 0 HREG
BIN 0 berepresented by C9B8 hex. Conversion of a hex number to itsdeci
mal equivalent follows the standar dcalculation.i.e. B8 hex (B hex)* 16+ (8)*
1 11*16+8184 Always remember that a hex number has abase of 16 dec
imal, a decimal number hasa base of 10 and a binary number a basebase 2.£ EX
AMPLES OF HEXADECIMAL NOTATION Now that hex notation has been intro-duced we
can discuss how the computerstores programs in more detail. Asalready
mentioned, the instructionsare stored memory as numbers. Someinstructio
ns only require a singlenumber, others require more.F Column 2 onthe Simul
ator display shows the memorycontents for the instructions in hex. All the nu
mbers used to describe aninstruction are given in sequence alongthe line. N
otice that numbers called upin an instruction appear towards the endof the strin
g of numbers. The earliernumbers determine the type ofinstruction.
Since the total number ofinstructions exceed 256, a prefixnumber is us
ed to produce other sets ofinstructions. The prefixes you will findare ED, CB,
DD and FD.£ Luckily assemblers takes account of allthese complexities, and yo
u will notneed to learn which numbers representwhich instructions. The Simul
ator can be switched betweendecimal and hexadecimal display bypressing s
hift "T", when a program isnot running.£ LD A,25H ADD A,A DAA LD HL,4589H
LD DE,3812H LD A,L SUB E DAA LD L,A LD A,H SBC A,D DAA LD H,A RET
BINARY CODED DECIMAL NOTATION We would still like to be able toperfor
m calculations on decimal numberswithout having to convert to binary orhexade
cimal form. Hexidecimal form holdsa key to a method of achieving this. The hex
digit requires adjustment toavoid the digits A-F. Adding 6 to thedigit, if
these symbols appear, resultsin correct adjustment. Take the hex number C
( 12 dec.).Adding 6 gives 18 decimal or 12 hex.Hence the hex number looks i
dentical tothe decimal answer required.£ Each digit is held in 4 bits of the
register. These 4 bits are called anibble, the whole 8 bit word is termed a
byte. Who said programmers have no senseof humour This presentation is termed
BinaryCoded Decimal or BCD. The Z80 accommodates the aboveadjustment
, with the instruction DAA,or Decimal Adjust Accumulator. Thisinstructio
n is used after an 8 bitADD, ADC, SUB, or SBC to adjust theAccumulato
r contents by adding 6 to thenibble, if either nibble in the answeris above 9
.£ To perform this operation the Z80 usestwo further flags, the H flag (C a H
alfcarry between nibbles ), and N flag(subtract instruction performed las
t).Since these flags cannot be tested aspart of a conditional jump etc. they
areof little use.£ EXAMPLES OF B.C.D. NOTATION These examples show simple ad
dition andrusubtraction in BCD format. Remember tokeep in Hex display mo
de or theillustrations will not appear in BCD.£ LD A,25H LD HL,STORE LD DE,
2468H INC (HL) RET BIN 65H BIN 87H BIN 9AH BIN DFH BIN FFH BIN 64H S
TORE DEFB 64 POSITIVE AND NEGATIVE NUMBER NOTATION So far we have only dea
lt with positivenumbers. If we had taken 6 from 5 wewould end up with the
answer 255 withcarry set. There is an interpretation ofnumbers that allows us
to consider thisanswer as the negative number -1. In this interpretation -1 m
ust equal255 or 11111111, or the processor wouldnot be able to take 6 from 5
and getthe right answer. Further 5-7 -2 or 254or 11111110. Bit 7, the most sig
nificantbit, represents the sign ( + oQr - ) ofthe number. When bit 7 is "1" t
he numberis negative, and when "0" the number ispositive.£ A useful operation
would be to make apositive number negative. Changing bit 7does not do this.
Take the positivenumber 2 or 00000010 and its negativeequivalent -2 or 111
11110 for example.Inverting all bits of the binary numberof +2 gives 111111
01, which is 1 lessthan that for -2.Hence to make a positive number negativewe
invert all its binary bits, known asComplimenting, and add 1. The nzhotationi
s generally termed 2s complement£i.e. 0 0 0 0 0 1 0 1 +5 invert
1 1 1 1 1 0 1 0 add 1 1 1 1 1 1 0 1 1 -5 invert 0 0 0 0 0 1 0 0 a
dd 1 0 0 0 0 0 1 0 1 +5 The operation luckily also makes anegative
number positive. A veryimportant point. The largest positive number we c
an holdin a single register, using thisnotation is 01111111 or 127
and thelargest negative number 10000000 or-128.£ The Z80 has two instruct
ions that maybe -used for these operations CPL complements or inverts t
he contents of the Accumulator. NEG negates, or makes negative, the
contents of the Accumulator by complementing and adding 1 in one
operation. A Sign flag FR(S) is provided on the Z80.It duplicates the sign
( bit 7 ) of theanswer after any arithmetic operation onthe Accumulator. The s
ign flag is "0"for a positive result and "1" for anegative result.£ When p
erforming arithmetic operationswhich we wish to interpret within therange
-128 to +127, the Carry flag nolonger signals an out of range result.However
another flag, the overflow (P/V)flag does. It is a "1" whenever theanswer
is outside the range -128 to +127and "0" within the range. It iseffecti
vely a carry into bit 7 whichwould make the sign bit incorrect.£ There is
no reason for sticking to 8bits using this method. So long as thenumber of b
its is sufficient and themost significant bit is taken as thesign bit a
ny size positive or negativenumber can be represented. Performing a similar calc
ulation to theabove we can show that a register paircan represent a number
in the range+32767 to -32768. The Sign and overflow flags are alsooperative
after AvDC and SBC instructionon the HL register. It reflects the 15thbit (or
bit 7 of H register) .£ It is important to remember that theZero, Carry, Si
gn, and overflow flagsare always operative after an 8 bitarithmetic instr
uction or a 16 bitaddition involving the Carry. Yourinterpretation of
the result determinesin which flag(s) you should beinterested in.£
EXAMPLES OF +VE AND -VE NOTATION Although these examples appear to beall po
sitive, they can be viewed equallyas nee%gative where appropriate and theSign
and overflow flags observedoperating. Remember a number above 12
8 isnegative. Since all the negative numbersin the examples are small, they
can beseen quickly and simply be converted bysubtractinSg the number from 256.
i.e. -2 is equivalent to 254 etc.£ LD A,254 NEG NEG ADD A,2 INC A LD HL,2
LD DE,FFFEH ADD HL,DE CPL LD HL,STORE LD (HL),7FH INC (HL) RET STORE B
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