2018-02-18 06:50:50 +00:00
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; disk2.asm
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;
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; This is the DISASSEMBLED source code for the Disk II controller ROM.
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; It adds up to 256 bytes of program code, which is all any peripheral
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; card was afforded.
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;
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; NOTE THAT THIS SOURCE CODE IS NOT ORIGINAL TO APPLE. I translated by
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; hand from the machine code in the ROM. Any comments, etc. you see
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; here, are from me--NOT APPLE.
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2018-02-18 17:16:09 +00:00
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;
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; For details on the assembly instructions--what they mean and do--this
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; is a good resource:
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; http://www.e-tradition.net/bytes/6502/6502_instruction_set.html
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;
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; Any number that has a $ in front of it means it's a hex number, vs.
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; decimal.
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2018-02-18 06:50:50 +00:00
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2018-02-18 17:16:09 +00:00
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; Our definitions for this little program. The EQU symbol is not a
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; formal instruction understood by the 6502 CPU; it's a notation that
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; simply means "equals"; e.g. GBASL equals $26.
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2018-02-18 06:50:50 +00:00
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GBASL EQU $26
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GBASH EQU $27
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BAS2H EQU $2B
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A1L EQU $3C
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A1H EQU $3D
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A3L EQU $40
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A3H EQU $41
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STACK EQU $0100
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GCRALT EQU $02D6
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XORSAV EQU $0300
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GCRTAB EQU $0356
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ENTRY EQU $0800
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PHASEOFF EQU $C080
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PHASEON EQU $C081
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TURNON EQU $C089
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SLCTD1 EQU $C08A
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READ EQU $C08C
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SETRD EQU $C08E
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WAIT EQU $FCA8
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FINDSLOT EQU $FF58
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; This is not needed by the disk controller program itself, but is used
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; by the system ROM to determine if there is a valid controller program
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; here. (There is!)
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00:A2 20 LDX #$20
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; Here we will write the group coded recording table for our decode
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; process, which is in the GCRTAB address.
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02:A0 00 LDY #$00
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04:A2 03 LDX #$03 ; our loop begins at $03
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06:86 3C NEXTGCR STX A1L ; A1L tracks the loop counter
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08:8A TXA
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09:0A ASL A
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0A:24 3C BIT A1L ; if X and X<<1 have no bits in common
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0C:F0 10 BEQ CONTGCR ; then X will not be written into GCRTAB
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; This sequence of operations will prime A for the VALIDENT check. Any
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; valid X register value (from which A is derived) will be one where
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; these three operations results in $40; after we loop on VALIDENT and
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; shift A to the right a bunch of times, we'll end up with $00 and exit
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; the loop without tripping the BCS (because none of the first 6 bits
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; were ever high).
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0E:05 3C ORA A1L
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10:49 FF EOR #$FF
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12:29 7E AND #$7E
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14:B0 08 VALIDENT BCS CONTGCR
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16:4A LSR A
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17:D0 FB BNE VALIDENT
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; Only certain X register values will be written via the STA
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; instruction; what we do write is an iteration of Y from $00..$3F.
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19:98 TYA
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1A:9D 56 03 STA GCRTAB,X
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1D:C8 INY
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1E:E8 CONTGCR INX
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1F:10 E5 BPL NEXTGCR
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; All this wrangling is here to make a record of the slot number.
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; Because JSR will push the calling address into the stack, we can find
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; the MSB of that address with the LDA STACK,X instruction. All of the
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; ASLs will essentially push the 7 one hex digit over, so $C7 becomes
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; $70. And we store that in BAS2H so we can use it to run operations on
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; the peripheral.
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21:20 58 FF JSR FINDSLOT
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24:BA TSX
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25:BD 00 01 LDA STACK,X ; this will load $C7 into A
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28:0A ASL
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29:0A ASL
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2A:0A ASL
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2B:0A ASL ; and now we have $70
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2C:85 2B STA BAS2H
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; Ok, with that done, we're going to get everything set up to copy the
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; zero track into RAM. NOTE: I'm not entirely sure why we're doing a
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; READ from the drive before we know we have drive 1 selected and turned
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; on.
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2E:AA TAX
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2F:BD 8E C0 LDA SETRD,X
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32:BD 8C C0 LDA READ,X
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35:BD 8A C0 LDA SLCTD1,X
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38:BD 89 C0 LDA TURNON,X
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; This loop is going to go through the stepper motor phases, flipping
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; them off and on again. To begin with, X is $70, so we're going to work
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; with phase 0 at the start.
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3B:A0 50 LDY #$50
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3D:BD 80 C0 PHASELOOP LDA PHASEOFF,X
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40:98 TYA
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41:29 03 AND #$03 ; drop all but the first 2 bits
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43:0A ASL ; and shift over
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44:05 2B ORA BAS2H ; and add that to $70
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46:AA TAX
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47:BD 81 C0 LDA PHASEON,X
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4A:A9 56 LDA #$56
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2018-02-19 01:49:34 +00:00
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; In at least one implementation (notably WinApple), the opcode below is
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; rewritten as `A9 00 EA`, which is equivalent to:
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; LDA #$00
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; NOP
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; This would essentially remove the WAIT call. The WAIT subroutine will,
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; in the course of its operation, leave $00 in A, which explains the LDA
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; #$00 opcode sequence. The NOP is there to replace the third byte
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; (which was part of the JSR address in its original form).
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2018-02-18 06:50:50 +00:00
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4C:20 A8 FC JSR WAIT ; wait for the motor
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4F:88 DEY
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50:10 EB BPL PHASELOOP
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; We're setting things up so we can start writing our decoded data into
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; the $08 page in memory, which is where we will ultimately jump to once
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; we finish going through track zero.
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52:85 26 STA GBASL ; A is $00 by this point
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54:85 3D STA A1H
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56:85 41 STA A3H
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58:A9 08 LDA #$08
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5A:85 27 STA GBASH ; so GBASH/L will hold $0800
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; We're going to check to see if we are at a header marker.
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5C:18 CHKHD CLC
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5D:08 CHKHDC PHP ; hang onto the status for later
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; Read byte from the disk (BPL is used here because anything that
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; doesn't have bit 7 high is bad data in 6-and-2 encoding).
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5E:BD 8C C0 READHD1 LDA READ,X
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61:10 FB BPL READHD1
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63:49 D5 CHKHD1 EOR #$D5
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65:D0 F7 BNE READHD1 ; try again
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; Look for the second header byte
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67:BD 8C C0 READHD2 LDA READ,X
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6A:10 FB BPL READHD2
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6C:C9 AA CHKHD2 CMP #$AA
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6E:D0 F3 BNE CHKHD1
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70:EA NOP ; I don't know why we NOP here
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; Third header byte
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71:BD 8C C0 READHD3 LDA READ,X
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74:10 FB BPL READHD3
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76:C9 96 CMP #$96 ; is this the end of a track marker?
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78:F0 09 BEQ METADATA ; seems to be!
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7A:28 PLP
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7B:90 DF BCC CHKHD ; if A < $96, keep seeking for a header byte
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7D:49 AD EOR #$AD ; if NOT, then this might be the end of a sector header
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7F:F0 25 BEQ DECODE ; so let's get decoding!
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81:D0 D9 BNE CHKHD ; Some other byte we didn't expect...
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; The metadata is 4-and-4 encoded, which are two bytes that are read in
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; sequence and then AND'd together. The second in the sequence is what
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; will stay behind in A3L; we'll read 3 sequences in all.
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83:A0 03 METADATA LDY #$03
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85:85 40 AGAIN44 STA A3L
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87:BD 8C C0 FIRST44 LDA READ,X ; read a byte
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8A:10 FB BPL FIRST44
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8C:2A ROL A
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8D:85 3C STA A1L
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8F:BD 8C C0 SECOND44 LDA READ,X ; read another byte
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92:10 FB BPL SECOND44
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94:25 3C AND A1L ; intersect with the shifted FIRST44
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96:88 DEY
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97:D0 EC BNE AGAIN44
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; This is going to pull from before we began checking for a header
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99:28 PLP
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9A:C5 3D CMP A1H
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9C:D0 BE BNE CHKHD
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; A3H can only be $00, and A3L will have been $96 from the last header
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; byte we read; since $96 - $00 will of course not be zero, this will
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; force a branch back to CHKHD. Why we have this code here is unclear to
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; me.
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9E:A5 40 LDA A3L
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A0:C5 41 CMP A3H
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A2:D0 B8 BNE CHKHD
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; If C is set, we will jump back to read the next header, _but_ we will
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; not execute the CLC instruction.
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A4:B0 B7 BCS CHKHDC
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; As we decode bytes, we're referencing the GCRTAB entries we built
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; earlier but from a slightly different address point (hence GCRALT).
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; But make no mistake--we're EORing with GCRTAB data. Note that XORSAV
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; is an entry point ($0300) which is conveniently(!) $56 less than
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; GCRTAB ($0356). It is, though, really just a place to stash the
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; intermediate data.
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A6:A0 56 DECODE LDY #$56 ; loop this many times...
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A8:84 3C SAV2BITS STY A1L ; save in A1L, because we use Y to read
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AA:BC 8C C0 DECBYTE2 LDY READ,X
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AD:10 FB BPL DECBYTE
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AF:59 D6 02 EOR GCRALT,Y
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B2:A4 3C LDY A1L
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B4:88 DEY ; decrement the loop counter
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B5:99 00 03 STA XORSAV,Y ; hang onto the EOR data
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B8:D0 EE BNE SAV2BITS
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; Looping from zero, now, we're going to write all that intermediate
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; data into the $0800 page (which is what (GBASL),Y resolves to),
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; counting up from $0800..$08FF.
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BA:84 3C SAV6BITS STY A1L
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BC:BC 8C C0 DECBYTE6 LDY READ,X
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BF:10 FB BPL DECBYTE6
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C1:59 D6 02 EOR GCRALT,Y
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C4:A4 3C LDY A1L
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C6:91 26 STA (GBASL),Y
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C8:C8 INY
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C9:D0 EF BNE SAV6BITS
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; We read ONE more byte, then determine if we need to check for another
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; header again.
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CB:BC 8C C0 FINBYTE LDY READ,X
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CE:10 FB BPL FINBYTE
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D0:59 D6 02 EOR XORTMP,Y
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; We may also get to here because it looks like we didn't write data
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; properly into the ENTRY page.
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D3:D0 87 BADDATA BNE CHKHD ; another sector?
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; We're using the first 89 ($56) bytes in XORSAV (which, remember, is
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; $56 less than the GCRTAB address point); if we go below $00 (rolling
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; over to $FF), start over.
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;
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; The bytes we've already written into the ENTRY page need to have those
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; 2 bits we compiled into those 89 bytes pushed back into the data.
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D5:A0 00 LDY #$00
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D7:A2 56 BITLOOP LDX #$56
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D9:CA WRITELOOP DEX
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DA:30 FB BMI BITLOOP ; start over if we went past $00
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DC:B1 26 LDA (GBASL),Y ; load $0800 + Y
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DE:5E 00 03 LSR XORSAV,X ; move bit 0 into carry
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E1:2A ROL A ; now load carry into A, plus the orig contents
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E2:5E 00 03 LSR XORSAV,X ; shift the former bit 1 (now bit 0) into carry again
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E5:2A ROL A ; and again load into A; now we have all 8 bits
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E6:91 26 STA (GBASL),Y ; and save it back to $0800 + Y
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E8:C8 INY
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E9:D0 EE BNE WRITELOOP ; we'll loop here 256 times
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; We're in the home stretch... we're just double-checking if we copied
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; things into the ENTRY page properly.
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2018-02-19 19:53:48 +00:00
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EB:E6 27 INC GBASH ; so now GBASL/H is $0900
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2018-02-18 06:50:50 +00:00
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ED:E6 3D INC A1H
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EF:A5 3D LDA A1H
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F1:CD 00 08 CMP ENTRY ; if A < ENTRY
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F4:A6 2B LDX BAS2H
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F6:90 DB BCC BADDATA ; then go back and try again
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F8:4C 01 08 JMP ENTRY+1 ; otherwise, let's boot the software!
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