; disk2.asm ; ; This is the DISASSEMBLED source code for the Disk II controller ROM. ; It adds up to 256 bytes of program code, which is all any peripheral ; card was afforded. ; ; NOTE THAT THIS SOURCE CODE IS NOT ORIGINAL TO APPLE. I translated by ; hand from the machine code in the ROM. Any comments, etc. you see ; here, are from me--NOT APPLE. ; ; For details on the assembly instructions--what they mean and do--this ; is a good resource: ; http://www.e-tradition.net/bytes/6502/6502_instruction_set.html ; ; Any number that has a $ in front of it means it's a hex number, vs. ; decimal. ; Our definitions for this little program. The EQU symbol is not a ; formal instruction understood by the 6502 CPU; it's a notation that ; simply means "equals"; e.g. GBASL equals $26. GBASL EQU $26 GBASH EQU $27 BAS2H EQU $2B A1L EQU $3C A1H EQU $3D A3L EQU $40 A3H EQU $41 STACK EQU $0100 GCRALT EQU $02D6 XORSAV EQU $0300 GCRTAB EQU $0356 ENTRY EQU $0800 PHASEOFF EQU $C080 PHASEON EQU $C081 TURNON EQU $C089 SLCTD1 EQU $C08A READ EQU $C08C SETRD EQU $C08E WAIT EQU $FCA8 FINDSLOT EQU $FF58 ; This is not needed by the disk controller program itself, but is used ; by the system ROM to determine if there is a valid controller program ; here. (There is!) 00:A2 20 LDX #$20 ; Here we will write the group coded recording table for our decode ; process, which is in the GCRTAB address. 02:A0 00 LDY #$00 04:A2 03 LDX #$03 ; our loop begins at $03 06:86 3C NEXTGCR STX A1L ; A1L tracks the loop counter 08:8A TXA 09:0A ASL A 0A:24 3C BIT A1L ; if X and X<<1 have no bits in common 0C:F0 10 BEQ CONTGCR ; then X will not be written into GCRTAB ; This sequence of operations will prime A for the VALIDENT check. Any ; valid X register value (from which A is derived) will be one where ; these three operations results in $40; after we loop on VALIDENT and ; shift A to the right a bunch of times, we'll end up with $00 and exit ; the loop without tripping the BCS (because none of the first 6 bits ; were ever high). 0E:05 3C ORA A1L 10:49 FF EOR #$FF 12:29 7E AND #$7E 14:B0 08 VALIDENT BCS CONTGCR 16:4A LSR A 17:D0 FB BNE VALIDENT ; Only certain X register values will be written via the STA ; instruction; what we do write is an iteration of Y from $00..$3F. 19:98 TYA 1A:9D 56 03 STA GCRTAB,X 1D:C8 INY 1E:E8 CONTGCR INX 1F:10 E5 BPL NEXTGCR ; All this wrangling is here to make a record of the slot number. ; Because JSR will push the calling address into the stack, we can find ; the MSB of that address with the LDA STACK,X instruction. All of the ; ASLs will essentially push the 7 one hex digit over, so $C7 becomes ; $70. And we store that in BAS2H so we can use it to run operations on ; the peripheral. 21:20 58 FF JSR FINDSLOT 24:BA TSX 25:BD 00 01 LDA STACK,X ; this will load $C7 into A 28:0A ASL 29:0A ASL 2A:0A ASL 2B:0A ASL ; and now we have $70 2C:85 2B STA BAS2H ; Ok, with that done, we're going to get everything set up to copy the ; zero track into RAM. NOTE: I'm not entirely sure why we're doing a ; READ from the drive before we know we have drive 1 selected and turned ; on. 2E:AA TAX 2F:BD 8E C0 LDA SETRD,X 32:BD 8C C0 LDA READ,X 35:BD 8A C0 LDA SLCTD1,X 38:BD 89 C0 LDA TURNON,X ; This loop is going to go through the stepper motor phases, flipping ; them off and on again. To begin with, X is $70, so we're going to work ; with phase 0 at the start. 3B:A0 50 LDY #$50 3D:BD 80 C0 PHASELOOP LDA PHASEOFF,X 40:98 TYA 41:29 03 AND #$03 ; drop all but the first 2 bits 43:0A ASL ; and shift over 44:05 2B ORA BAS2H ; and add that to $70 46:AA TAX 47:BD 81 C0 LDA PHASEON,X 4A:A9 56 LDA #$56 ; In at least one implementation (notably WinApple), the opcode below is ; rewritten as `A9 00 EA`, which is equivalent to: ; LDA #$00 ; NOP ; This would essentially remove the WAIT call. The WAIT subroutine will, ; in the course of its operation, leave $00 in A, which explains the LDA ; #$00 opcode sequence. The NOP is there to replace the third byte ; (which was part of the JSR address in its original form). 4C:20 A8 FC JSR WAIT ; wait for the motor 4F:88 DEY 50:10 EB BPL PHASELOOP ; We're setting things up so we can start writing our decoded data into ; the $08 page in memory, which is where we will ultimately jump to once ; we finish going through track zero. 52:85 26 STA GBASL ; A is $00 by this point 54:85 3D STA A1H 56:85 41 STA A3H 58:A9 08 LDA #$08 5A:85 27 STA GBASH ; so GBASH/L will hold $0800 ; We're going to check to see if we are at a header marker. 5C:18 CHKHD CLC 5D:08 CHKHDC PHP ; hang onto the status for later ; Read byte from the disk (BPL is used here because anything that ; doesn't have bit 7 high is bad data in 6-and-2 encoding). 5E:BD 8C C0 READHD1 LDA READ,X 61:10 FB BPL READHD1 63:49 D5 CHKHD1 EOR #$D5 65:D0 F7 BNE READHD1 ; try again ; Look for the second header byte 67:BD 8C C0 READHD2 LDA READ,X 6A:10 FB BPL READHD2 6C:C9 AA CHKHD2 CMP #$AA 6E:D0 F3 BNE CHKHD1 70:EA NOP ; I don't know why we NOP here ; Third header byte 71:BD 8C C0 READHD3 LDA READ,X 74:10 FB BPL READHD3 76:C9 96 CMP #$96 ; is this the end of a track marker? 78:F0 09 BEQ METADATA ; seems to be! 7A:28 PLP 7B:90 DF BCC CHKHD ; if A < $96, keep seeking for a header byte 7D:49 AD EOR #$AD ; if NOT, then this might be the end of a sector header 7F:F0 25 BEQ DECODE ; so let's get decoding! 81:D0 D9 BNE CHKHD ; Some other byte we didn't expect... ; The metadata is 4-and-4 encoded, which are two bytes that are read in ; sequence and then AND'd together. The second in the sequence is what ; will stay behind in A3L; we'll read 3 sequences in all. 83:A0 03 METADATA LDY #$03 85:85 40 AGAIN44 STA A3L 87:BD 8C C0 FIRST44 LDA READ,X ; read a byte 8A:10 FB BPL FIRST44 8C:2A ROL A 8D:85 3C STA A1L 8F:BD 8C C0 SECOND44 LDA READ,X ; read another byte 92:10 FB BPL SECOND44 94:25 3C AND A1L ; intersect with the shifted FIRST44 96:88 DEY 97:D0 EC BNE AGAIN44 ; This is going to pull from before we began checking for a header 99:28 PLP 9A:C5 3D CMP A1H 9C:D0 BE BNE CHKHD ; A3H can only be $00, and A3L will have been $96 from the last header ; byte we read; since $96 - $00 will of course not be zero, this will ; force a branch back to CHKHD. Why we have this code here is unclear to ; me. 9E:A5 40 LDA A3L A0:C5 41 CMP A3H A2:D0 B8 BNE CHKHD ; If C is set, we will jump back to read the next header, _but_ we will ; not execute the CLC instruction. A4:B0 B7 BCS CHKHDC ; As we decode bytes, we're referencing the GCRTAB entries we built ; earlier but from a slightly different address point (hence GCRALT). ; But make no mistake--we're EORing with GCRTAB data. Note that XORSAV ; is an entry point ($0300) which is conveniently(!) $56 less than ; GCRTAB ($0356). It is, though, really just a place to stash the ; intermediate data. A6:A0 56 DECODE LDY #$56 ; loop this many times... A8:84 3C SAV2BITS STY A1L ; save in A1L, because we use Y to read AA:BC 8C C0 DECBYTE2 LDY READ,X AD:10 FB BPL DECBYTE AF:59 D6 02 EOR GCRALT,Y B2:A4 3C LDY A1L B4:88 DEY ; decrement the loop counter B5:99 00 03 STA XORSAV,Y ; hang onto the EOR data B8:D0 EE BNE SAV2BITS ; Looping from zero, now, we're going to write all that intermediate ; data into the $0800 page (which is what (GBASL),Y resolves to), ; counting up from $0800..$08FF. BA:84 3C SAV6BITS STY A1L BC:BC 8C C0 DECBYTE6 LDY READ,X BF:10 FB BPL DECBYTE6 C1:59 D6 02 EOR GCRALT,Y C4:A4 3C LDY A1L C6:91 26 STA (GBASL),Y C8:C8 INY C9:D0 EF BNE SAV6BITS ; We read ONE more byte, then determine if we need to check for another ; header again. CB:BC 8C C0 FINBYTE LDY READ,X CE:10 FB BPL FINBYTE D0:59 D6 02 EOR XORTMP,Y ; We may also get to here because it looks like we didn't write data ; properly into the ENTRY page. D3:D0 87 BADDATA BNE CHKHD ; another sector? ; We're using the first 89 ($56) bytes in XORSAV (which, remember, is ; $56 less than the GCRTAB address point); if we go below $00 (rolling ; over to $FF), start over. ; ; The bytes we've already written into the ENTRY page need to have those ; 2 bits we compiled into those 89 bytes pushed back into the data. D5:A0 00 LDY #$00 D7:A2 56 BITLOOP LDX #$56 D9:CA WRITELOOP DEX DA:30 FB BMI BITLOOP ; start over if we went past $00 DC:B1 26 LDA (GBASL),Y ; load $0800 + Y DE:5E 00 03 LSR XORSAV,X ; move bit 0 into carry E1:2A ROL A ; now load carry into A, plus the orig contents E2:5E 00 03 LSR XORSAV,X ; shift the former bit 1 (now bit 0) into carry again E5:2A ROL A ; and again load into A; now we have all 8 bits E6:91 26 STA (GBASL),Y ; and save it back to $0800 + Y E8:C8 INY E9:D0 EE BNE WRITELOOP ; we'll loop here 256 times ; We're in the home stretch... we're just double-checking if we copied ; things into the ENTRY page properly. EB:E6 27 INC GBASH ; so now GBASL/H is $0900 ED:E6 3D INC A1H EF:A5 3D LDA A1H F1:CD 00 08 CMP ENTRY ; if A < ENTRY F4:A6 2B LDX BAS2H F6:90 DB BCC BADDATA ; then go back and try again F8:4C 01 08 JMP ENTRY+1 ; otherwise, let's boot the software!