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5fd995aa626c0a59a0d4c99152ca7a6ed1f5220d
SplitBit-Emulator/Programs/CosmOS/Source/cosmos.asm
T
AnachronautandClaude Opus 5 5fd995aa62 Breakpoints: SWI osBreak, and s refuses a read only bank 
A breakpoint that shows every register as the program had them, waits for a
key, and carries on.

NOTHING IS OVERWRITTEN, and that is the design rather than a shortcut. A
breakpoint poked into a running program has to replace an instruction, and
putting that instruction back in order to continue is the same act as
disarming the breakpoint; firing a second time would mean stepping over the
restored instruction and putting the breakpoint back behind it, and this
machine cannot step a single instruction. SWI is two bytes, dispatches
through a vector, and its frame already holds the address after it, so RETI
resumes at the next instruction with nothing to restore and nothing to
re-arm. It fires every time it is reached.

The price is that a breakpoint is part of the program: a build with them in
has different addresses from a build without. That is the bargain every
machine with a break instruction makes.

Every value shown comes out of the frame rather than the registers, because
by the time the handler runs the registers are the handler's. Apps/Break.asm
stops twice so that the second stop is checked as well as the first.

Also here, found by the test that came with it: the monitor's s wrote into
whichever bank was selected, and bank 2 is the controller's own table,
published read only. Writing to it was refused, and a refusal nobody catches
stops the machine - so selecting the bank table to look at it and then typing
s killed the session. bankPresent now keeps the whole flags byte and s
declines. The recorded output of cosmosMonitor had contained that crash,
having been blessed without being read.

Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
2026-08-19 22:09:21 -04:00

2174 lines
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; cosmos.asm
; CosmOS, and the shell that is most of it.
;
; The machine boots into this. It registers what the hardware brought, mounts whatever
; disk is attached, and then reads lines and does what they say until there is no more
; typing to be had.
;
; ---- Where things live ----
;
; The system keeps to the bottom of both memories, and everything above is for whatever
; it is running:
;
; Program Memory 0x0000 - 0x1FFF the system
; 0x2000 - a loaded program's code
; Data Memory 0x0000 - 0x0FFF the system
; 0x1000 - a loaded program's data
;
; Nothing enforces that. Nothing can: the fence guards a range, and this is a convention
; about which range belongs to whom rather than a rule about what may be touched. The
; assembler prints both segment sizes, and they are what to watch.
;
; A program is staged at 0x8000 while it is being loaded, which is inside the region a
; loaded program will own. That is safe because nothing is running during a load, and it
; is where a big program can be read without the system reserving the room for good.
;
; ---- What it can do ----
;
; dir List what is on the disk.
; load Read a program off the disk and put it where it asks to go.
; run Start the program that was loaded.
; help Say what these are.
; exit Stop.
;
; dump is next. The dispatch below is a chain of comparisons, which is the right shape for
; five commands and the wrong shape for twenty; when it grows, the table that
; dispatchTest.asm demonstrates is where it should go.
;
; Written by Anachronaut
#Include console.asm
#Include text.asm
#Include sbfs.asm
#Include services.asm
#Program
boot:
SETD.0 Banner
CALL printString
CALL newLine
; Find out whether there is a filesystem to talk to. Doing this once at boot rather than
; once per command means a disk swapped underneath us is not noticed, which is honest
; for a machine whose disk is a file named on the command line.
CALL sbfsMount
SETD.0 DiskReady
BNQ bootNoDisk
INIA 0x01
STA.0
BRI prompt
bootNoDisk:
RSTA
STA.0
SETD.0 NoDisk
CALL printString
CALL newLine
; ---- The loop ----
; ---- The loop ----
;
; The shell has two modes and one prompt that says which. Ordinary mode runs programs;
; monitor mode also looks at memory, changes it, and jumps into it.
;
; THE MODE IS A VARIABLE RATHER THAN A SECOND LOOP, and that is what makes it persistent
; without anything having to remember it. Every way back here goes through this one place,
; INCLUDING A PROGRAM GIVING THE MACHINE BACK - so jumping to an address, letting it run,
; and having it exit puts you back at the monitor prompt you left from, rather than at the
; shell. Only saying so leaves the monitor, or a program breaking the machine badly enough
; to need starting again.
prompt:
SETD.0 Mode
LDA.0
BRA promptPlain
SETD.0 MonitorPrompt
BRI promptSay
promptPlain:
SETD.0 PromptText
promptSay:
CALL printString
SETD.0 CommandLine
INIB 0d63
CALL readLine
; Running out of typing is how this ends. It is not the same as an empty line, which is
; just somebody pressing return, and the shell should sit there when that happens.
SETD.0 ConsoleEndOfInput
LDA.0
BNA quitRanOut
SETD.0 CommandLine
CALL textSplit
; An empty line asks for nothing.
SETD.0 CommandLine
LDA.0
BRA prompt
SETD.0 CommandLine
SETD.1 DirName
CALL textSame
BRQ doDir
SETD.0 CommandLine
SETD.1 LoadName
CALL textSame
BRQ doLoad
SETD.0 CommandLine
SETD.1 RunName
CALL textSame
BRQ doRun
SETD.0 CommandLine
SETD.1 DeleteName
CALL textSame
BRQ doDelete
SETD.0 CommandLine
SETD.1 RenameName
CALL textSame
BRQ doRename
SETD.0 CommandLine
SETD.1 HelpName
CALL textSame
BRQ doHelp
SETD.0 CommandLine
SETD.1 ExitName
CALL textSame
BRQ doExit
SETD.0 CommandLine
SETD.1 MonitorName
CALL textSame
BRQ doMonitor
; The monitor's own commands, which only answer when the monitor is on. They are single
; letters because they are typed constantly and because the prompt has already said which
; mode you are in; the plain shell keeps its words and stays plain.
SETD.0 Mode
LDA.0
BRA promptUnknown
SETD.0 CommandLine
SETD.1 ExamineName
CALL textSame
BRQ doExamine
SETD.0 CommandLine
SETD.1 DisName
CALL textSame
BRQ doDisassemble
SETD.0 CommandLine
SETD.1 SetName
CALL textSame
BRQ doSet
SETD.0 CommandLine
SETD.1 BankName
CALL textSame
BRQ doBank
SETD.0 CommandLine
SETD.1 GoName
CALL textSame
BRQ doGo
promptUnknown:
; Nothing matched. Saying which word was not understood is worth the four instructions:
; it tells somebody who mistyped what they actually typed.
SETD.0 Unknown
CALL printString
SETD.0 CommandLine
CALL printString
CALL newLine
BRI prompt
; Running out of console leaves the cursor part way along a line, because there was no
; return at the end to move it on. Somebody who typed "exit" has already pressed one, and
; a second would only leave a blank line behind.
; Leaving whatever you are in: the monitor if you are in it, and the machine if you are
; not. Two exits to stop from the monitor, which is what every nested prompt has ever asked
; for and reads the right way round.
doExit:
SETD.0 Mode
LDA.0
BRA quit
RSTA
STA.0
BRI prompt
doMonitor:
INIA 0x01
SETD.0 Mode
STA.0
SETD.0 MonitorHelp
CALL printString
CALL newLine
BRI prompt
quitRanOut:
CALL newLine
quit:
SETD.0 Farewell
CALL printString
CALL newLine
HALT
; ---- dir ----
;
; Walks the directory and prints what is in it. A free entry in the middle of a directory
; is stepped over by the walk, so what comes out is the files and nothing else.
doDir:
SETD.0 DiskReady
LDA.0
BRA dirNoDisk
RSTA
SETD.0 DirSeen
STA.0
CALL sbfsFirst
BRI dirCheck
dirStep:
CALL sbfsNext
dirCheck:
BNQ dirDone
SETD.0 DirSeen
LDA.0
INCA
STA.0
SETD.0 SbfsName
CALL printString
SETD.0 SbfsName
CALL nameWidth
MVQA
CALL printSpaces
; A file's length is its block count times 256 plus its tail, which is the block count
; in the high byte and the tail in the low one. Nothing has to multiply anything.
SETD.0 SbfsFileBlocks
DPUP.0 0d01
LDA.0
SETD.1 DirSize
STA.1
SETD.0 SbfsFileTail
LDA.0
SETD.1 DirSize
INCD.1
STA.1
SETD.0 DirSize
CALL printWordDecimal
CALL newLine
BRI dirStep
dirDone:
SETD.0 DirSeen
LDA.0
CALL printByteDecimal
; One file is not one files. Cheap to get right and it reads as carelessness otherwise.
SETD.0 DirSeen
LDA.0
DECA
BRA dirOne
SETD.0 FilesText
BRI dirCount
dirOne:
SETD.0 FileText
dirCount:
CALL printString
CALL newLine
BRI prompt
dirNoDisk:
SETD.0 NoDisk
CALL printString
CALL newLine
BRI prompt
; DP0 names a string. Q is how many spaces pad it out to twenty four columns. A name
; already that long gets one space, so that it cannot run into the number after it.
nameWidth:
INIA 0d24
SETD.1 WidthLeft
STA.1
widthLoop:
LDA.0
BRA widthDone
SETD.1 WidthLeft
LDA.1
DECA
STA.1
BRA widthFloor
INCD.0
BRI widthLoop
widthFloor:
INIA 0d1
SETD.1 WidthLeft
STA.1
widthDone:
SETD.1 WidthLeft
LDA.1
RSTB
CCF
ADD
RET
; ---- load ----
;
; Reads a program off the disk and puts it where its header asks to go. Nothing relocates
; anything: the addresses in the header are the ones the program was built for, and it
; would not work anywhere else.
;
; The whole file is staged at 0x8000 first and then blitted into place, because where the
; pieces belong is not known until the header has been read, and the header is in the file.
doLoad:
SETD.0 DiskReady
LDA.0
BRA loadNoDisk
SETD.1 TextRest
LDD.0.1
LDA.0
BRA loadNothingNamed
CALL sbfsFind
BNQ loadMissing
SETD.1 0x80 0x00
CALL sbfsRead
BNQ loadUnreadable
; "SBEX", or this is not a program. Without this, loading a text file would put nonsense
; into Program Memory and then jump into the middle of it.
SETD.0 0x80 0x00
SETD.2 ExecMagic
INIA 0d4
SETD.1 LoadCount
STA.1
loadMagicLoop:
LDA.0
LDB.2
XOR
BNQ loadNotProgram
INCD.0
INCD.2
LDA.1
DECA
STA.1
BNA loadMagicLoop
; Version one is code and data. Version two also brings vectors, which is a thing a
; loader has to know how to do rather than a detail it can skip: a program whose handlers
; were quietly dropped would run and then go wrong somewhere with nothing to connect it
; back to here. Anything else is refused.
SETD.0 0x80 0x00
DPUP.0 0d04
LDA.0
SETD.1 LoadVersion
STA.1
INIB 0d1
XOR
BRQ loadVersionKnown
SETD.1 LoadVersion
LDA.1
INIB 0d2
XOR
BNQ loadWrongVersion
loadVersionKnown:
; The code. It comes from the staging area just past the sixteen byte header, and goes
; wherever the header says, in Program Memory, which the instruction set cannot write
; and the controller can.
INIA 0d1
OUTA 0xE0 ; SourceBank: Data Memory, where the file was staged.
INIA 0x80
OUTA 0xE1
INIA 0d16
OUTA 0xE2 ; 0x8010, the first byte after the header.
RSTA
OUTA 0xE3 ; DestBank: Program Memory.
SETD.0 0x80 0x00
DPUP.0 0d06
LDA.0
OUTA 0xE4
INCD.0
LDA.0
OUTA 0xE5
SETD.0 0x80 0x00
DPUP.0 0d10
LDA.0
OUTA 0xE6
INCD.0
LDA.0
OUTA 0xE7
INIA 0x01
OUTA 0xE8 ; Blit.
; Then the data. A blit leaves its addresses past whatever it touched, so the source is
; already sitting on the first byte of the data and only the destination changes.
INIA 0d1
OUTA 0xE3 ; DestBank: Data Memory.
SETD.0 0x80 0x00
DPUP.0 0d12
LDA.0
OUTA 0xE4
INCD.0
LDA.0
OUTA 0xE5
SETD.0 0x80 0x00
DPUP.0 0d14
LDA.0
OUTA 0xE6
INCD.0
LDA.0
OUTA 0xE7
INIA 0x01
OUTA 0xE8 ; Blit.
; ---- The vectors it brought ----
;
; Kept here rather than installed. A vector points into a program, so it has no business
; being in the table while that program is only loaded and not running: run puts them in
; and exit takes them out again, so the window they are live in is exactly the run.
; Keeping our own copy is also what lets a program be run more than once, since the
; staging area it came in on is fair game for the program's own use.
;
; Where to read them from is not worked out. The data blit left the controller's source
; address on the first byte after the data, which is where they are, so it is read back.
SETD.0 VectorSource
INA 0xE1
STA.0
INCD.0
INA 0xE2
STA.0
SETD.0 0x80 0x00
DPUP.0 0d05
LDA.0
SETD.1 LoadedVectorCount
STA.1
BRA loadVectorsCopied
; More than there is room for is refused rather than half taken. Half a program's
; handlers is not a smaller version of that program.
INIB 0d17
CCF
SUB
BNC loadTooManyVectors
SETD.0 VectorSource
LDD.2.0 ; DP2 walks the entries where they are staged.
SETD.3 LoadedVectors ; DP3 walks our own copy of them.
SETD.1 LoadedVectorCount
LDA.1
SETD.1 VectorsLeft
STA.1
loadVectorCopy:
; Four bytes: where it goes, then what goes there. The two bytes for what was there
; before are left alone until something is actually put in.
LDA.2
STA.3
INCD.2
INCD.3
LDA.2
STA.3
INCD.2
INCD.3
LDA.2
STA.3
INCD.2
INCD.3
LDA.2
STA.3
INCD.2
INCD.3
INCD.3
INCD.3
SETD.1 VectorsLeft
LDA.1
DECA
STA.1
BNA loadVectorCopy
loadVectorsCopied:
; Where it starts. Written out by hand rather than through a routine, because a routine
; could not hand two bytes back: CALL puts A, B and the first three pointers back the
; way it found them.
SETD.0 0x80 0x00
DPUP.0 0d08
LDA.0
SETD.1 LoadedEntry
STA.1
INCD.0
INCD.1
LDA.0
STA.1
INIA 0x01
SETD.0 LoadedOk
STA.0
SETD.0 LoadedText
CALL printString
SETD.0 LoadedEntry
CALL printWordHex
CALL newLine
BRI prompt
loadNoDisk:
SETD.0 NoDisk
BRI loadComplain
loadNothingNamed:
SETD.0 LoadWhat
BRI loadComplain
loadTooManyVectors:
SETD.0 TooManyVectors
BRI loadComplain
loadMissing:
SETD.0 NoSuchFile
BRI loadComplain
loadUnreadable:
SETD.0 Unreadable
BRI loadComplain
loadNotProgram:
SETD.0 NotProgram
BRI loadComplain
loadWrongVersion:
SETD.0 WrongVersion
loadComplain:
CALL printString
CALL newLine
BRI prompt
; ---- delete and rename ----
;
; The two things a disk needs that reading and writing do not provide, and the two that
; anything editing a document will want from the shell as well as from a program. Deleting
; frees an entry and its blocks; renaming changes twenty two bytes and moves nothing.
doDelete:
SETD.0 DiskReady
LDA.0
BRA fileNoDisk
SETD.1 TextRest
LDD.0.1
LDA.0
BRA deleteWhat
CALL sbfsDelete
BNQ deleteFailed
SETD.0 Deleted
CALL printString
CALL newLine
BRI prompt
deleteWhat:
SETD.0 DeleteWhat
BRI fileComplain
deleteFailed:
SETD.0 NoSuchFile
BRI fileComplain
doRename:
SETD.0 DiskReady
LDA.0
BRA fileNoDisk
SETD.1 TextRest
LDD.0.1
LDA.0
BRA renameWhat
; Two names, so the rest of the line is split again. textSplit writes a zero over the
; space it cuts at, so what was one string becomes two without anything being copied.
SETD.1 TextRest
LDD.0.1
SETD.1 RenameFrom
STD.0.1
CALL textSplit
SETD.1 TextRest
LDD.1.1
LDA.1
BRA renameWhat ; Only one name was given, and this needs both.
SETD.2 RenameFrom
LDD.0.2
CALL sbfsRename
BNQ renameFailed
SETD.0 Renamed
CALL printString
CALL newLine
BRI prompt
renameWhat:
SETD.0 RenameWhat
BRI fileComplain
renameFailed:
; Either there is no such file or the new name is already taken. Which of the two is not
; worth another message: both mean the disk does not have room for that name to move.
SETD.0 RenameNo
BRI fileComplain
fileNoDisk:
SETD.0 NoDisk
fileComplain:
CALL printString
CALL newLine
BRI prompt
; ---- run ----
;
; Hands the machine to whatever was loaded. Where the Stack is now is written down first,
; because the program is not going to unwind anything it pushes and the exit handler has
; to be able to put the Stack back.
doRun:
SETD.0 LoadedOk
LDA.0
BRA runNothing
MVSD.0
SETD.1 SystemStack
STD.0.1
; Whatever followed the word "run" is kept where the program can ask for it. Copied
; rather than pointed at, because what it is pointing at is the line the shell typed
; into, and a program is entitled to outlive the shell's opinion of that.
SETD.1 TextRest
LDD.0.1
SETD.1 RunArgument
INIB 0d64
CALL copyText
CALL installVectors
; The entry address is a number until BRD makes it a place. DP3 is the one to build it
; in, because it is the pointer nothing puts back.
SETD.1 LoadedEntry
LDD.3.1
BRD.3
runNothing:
SETD.0 NothingLoaded
CALL printString
CALL newLine
BRI prompt
; DP0 is a string, DP1 is where it should go, and B is how much room there is counting
; the zero on the end. What does not fit is left behind, and what is written is a string
; either way.
copyText:
BRB copyTextDone ; No room at all, so nothing is written, not even the zero.
copyTextLoop:
DECB
BRB copyTextEnd ; Only room for the terminator now.
LDA.0
STA.1
BRA copyTextDone
INCD.0
INCD.1
BRI copyTextLoop
copyTextEnd:
RSTA
STA.1
copyTextDone:
RET
; ---- Putting a program's vectors in, and taking them out again ----
;
; The vector table lives in Program Memory, which no instruction can write, so both of
; these go through the memory controller. Port 0xE9 reads a byte from the source and writes
; a byte to the destination, stepping the address on either way, so a two byte entry is two
; reads or two writes and no address arithmetic in between.
;
; What was in the slot is kept before anything replaces it, and put back afterwards, rather
; than the slot being cleared. Clearing would be wrong wherever a program has installed a
; handler over one the system was already using: the program is allowed to do that, and
; when it goes, what it covered up has to come back rather than becoming a hole.
installVectors:
SETD.0 LoadedVectorCount
LDA.0
BRA installDone
SETD.1 VectorsLeft
STA.1
SETD.3 LoadedVectors
installOne:
; DP3 walks one six byte entry: where it goes, what goes there, and room for what was
; there before. Reading and writing the same slot, so the controller is pointed at it
; from both ends at once and the address is only worked out once.
RSTA
OUTA 0xE0 ; SourceBank: Program Memory.
OUTA 0xE3 ; DestBank: the same.
LDA.3
OUTA 0xE1
OUTA 0xE4
INCD.3
LDA.3
OUTA 0xE2
OUTA 0xE5
INCD.3 ; On the handler.
; What is there now, before anything replaces it.
INA 0xE9
PSHA
INA 0xE9
PSHA
; And the handler in its place.
LDA.3
OUTA 0xE9
INCD.3
LDA.3
OUTA 0xE9
INCD.3 ; On the two bytes kept for what was there before.
; The Stack gives them back in the reverse of the order they went on, so the low byte
; arrives first and is written to the second of the two. Getting this the natural way
; round instead put the low byte where the high one goes and the high byte over the
; handler, which the first run of a program survives - the table is already written by
; then - and the second run does not.
POPA
INCD.3
STA.3
DECD.3
POPA
STA.3
INCD.3
INCD.3
SETD.1 VectorsLeft
LDA.1
DECA
STA.1
BNA installOne
installDone:
RET
removeVectors:
SETD.0 LoadedVectorCount
LDA.0
BRA removeDone
SETD.1 VectorsLeft
STA.1
SETD.3 LoadedVectors
removeOne:
RSTA
OUTA 0xE3 ; DestBank: Program Memory.
LDA.3
OUTA 0xE4
INCD.3
LDA.3
OUTA 0xE5
INCD.3
INCD.3
INCD.3 ; Past the handler, to what was underneath it.
LDA.3
OUTA 0xE9
INCD.3
LDA.3
OUTA 0xE9
INCD.3
SETD.1 VectorsLeft
LDA.1
DECA
STA.1
BNA removeOne
removeDone:
RET
; ---- The services ----
;
; These are what a loaded program is allowed to ask for. The names and their numbers come
; from services.asm, which the programs include as well, so neither side writes a number
; down and the two cannot disagree about them.
;
; A handler arrives with the caller's registers exactly as they were: an interrupt frame
; is pushed, not cleared. So the pointer a program put in DP0 is still there to be used.
; The disk finishing, acknowledged and ignored.
;
; The system drives the disk by asking its status port and waiting, so it has no use for
; the line. But the disk raises one after every operation whether anybody wants it or not,
; and a line goes on waiting while the Interrupt Flag is down rather than being lost. The
; shell keeps the flag down, so the line from the last disk read was still standing when
; the first program to enable interrupts ran, and it arrived there - a fault, in a program
; that had never heard of the disk, blamed on the innocent instruction that let it through.
;
; Answering a line is what takes it down, so this is one instruction and that is the point.
diskDone:
RETI
handlePrintString:
CALL printString
RETI
handleReadLine:
CALL readLine
; readLine works out how long the line was, and RETI would throw that away: it restores
; every register from the frame, which is exactly what makes an interrupt safe to arrive
; unannounced and exactly what stops a service answering. So the answer is written into
; the frame, over the saved Q, and RETI puts it back as though the caller had computed it.
;
; This has to be here rather than in a routine, because the offset is from where the
; Stack Pointer is now and a CALL moves it by ten.
MVQA
MVSD.1
DPUP.1 0d02
STA.1
RETI
; What the program was asked to work on. DP0 says where to put it and B how much room
; there is, counting the zero on the end, which is the same bargain readLine offers.
;
; Being asked for rather than left at an agreed address is deliberate. The two sides of
; this already have to agree on a vector number and nothing else, and that number is
; written down once in services.asm; an address would be a second thing to agree about, in
; a memory map that is a convention rather than anything enforced.
handleArgument:
PSHD.0
POPD.1
SETD.0 RunArgument
CALL copyText
RETI
; ---- The disk, on a program's behalf ----
;
; A loaded program that wanted a file used to include the whole filesystem, so it carried a
; private copy of code the system already has running, and mounted a disk that was already
; mounted. These are that code, reachable through a number instead.
;
; Every one of them answers in Q, and the answer is written into the frame over the saved
; register, because RETI puts every register back and would otherwise throw it away. That
; has to be done here rather than in a routine of its own: the offsets are from where the
; Stack Pointer is, and a CALL moves it by ten.
;
; A machine with no disk answers no to all of them rather than going ahead and finding out,
; because sbfs on a disk that was never mounted is reading whatever bank 3 happens to be.
; DP0 names the file, DP1 says where to put it. Q is zero if it read, and DP3 comes back
; holding how many bytes there were.
handleFileRead:
SETD.2 DiskReady
LDA.2
BRA fileReadNo
CALL sbfsFind
BNQ fileReadNo
; A file of 256 blocks is 64K, which will not fit in Data Memory and will not fit in the
; pointer that says how long it is either. Refused, rather than read as much of as fits:
; a length that lies is worse than a file that will not open.
SETD.2 SbfsFileBlocks
LDA.2
BNA fileReadNo
CALL sbfsRead
BNQ fileReadNo
; How long it is: the block count is the high byte of that and the tail is the low one,
; which is how a size is put together everywhere on this disk.
SETD.2 SbfsFileBlocks
INCD.2
LDA.2
SETD.2 SbfsFileTail
LDB.2
MVSD.2
DPUP.2 0d05 ; The saved DP3, high byte first.
STA.2
INCD.2
STB.2
MVSD.2
DPUP.2 0d02 ; And the saved Q.
RSTA
STA.2
RETI
fileReadNo:
MVSD.2
DPUP.2 0d02
INIA 0d1
STA.2
RETI
; DP0 names the file, DP1 is the bytes, and A and B together are how many. Q is zero if it
; saved. Whether it was there before makes no difference, which is what saving means.
handleFileSave:
SETD.2 DiskReady
PSHA
LDA.2
BRA fileSaveNoDisk
POPA
; Blocks are the high half of the count and the tail is the low half.
SETD.2 SbfsFileBlocks
PSHA
RSTA
STA.2 ; A whole file's block count fits in a byte, so this is zero.
INCD.2
POPA
STA.2
SETD.2 SbfsFileTail
STB.2
CALL sbfsSaveFile
MVQA
MVSD.2
DPUP.2 0d02
STA.2
RETI
fileSaveNoDisk:
POPA
MVSD.2
DPUP.2 0d02
INIA 0d1
STA.2
RETI
; DP0 names it. Q is zero if it went.
handleFileDelete:
SETD.2 DiskReady
LDA.2
BRA serviceNoDisk
CALL sbfsDelete
MVQA
MVSD.2
DPUP.2 0d02
STA.2
RETI
; DP0 is the name it has, DP1 the name it should have. Q is zero if it moved.
handleFileRename:
SETD.2 DiskReady
LDA.2
BRA serviceNoDisk
CALL sbfsRename
MVQA
MVSD.2
DPUP.2 0d02
STA.2
RETI
serviceNoDisk:
MVSD.2
DPUP.2 0d02
INIA 0d1
STA.2
RETI
; A breakpoint. Shows every register as the interrupted program had them, waits for a key,
; and returns as though nothing happened.
;
; EVERY VALUE COMES OUT OF THE FRAME, not out of the registers, because by the time this
; runs the registers belong to the handler. The frame is what the program had, and RETI is
; going to give it all back, so what is shown is what will be resumed with.
;
; DP3 holds the frame throughout. It survives a CALL, and console.asm promises not to
; disturb it, which is what lets the printing routines be used between one field and the
; next. The Stack Pointer comes back to the same place after a balanced call, so the frame
; stays where it was found.
;
; +1 Status +2 Q +3 A +4 B +5 DP3 +7 DP2 +9 DP1 +11 DP0 +13 where it resumes
handleBreak:
MVSD.3
; Where it broke, which is two before where it resumes: the SWI and the vector it names.
SETD.0 BreakText
CALL printString
PSHD.3
POPD.0
DPUP.0 0d13
LDA.0
PSHA ; The high half, while the low one is worked on.
INCD.0
LDA.0
INIB 0d2
CCF
SUB ; Two back from where it resumes: the SWI and the vector it names.
MVQA
POPB
BRC breakBorrowed ; It borrowed, so the high half comes down by one.
BRI breakAddress
breakBorrowed:
DECB
breakAddress:
PSHA ; low
PSHB ; high
POPA
CALL printByteHex
POPA
CALL printByteHex
CALL newLine
SETD.0 ARegText
PSHD.3
POPD.1
DPUP.1 0d3
CALL breakByte
SETD.0 BRegText
PSHD.3
POPD.1
DPUP.1 0d4
CALL breakByte
SETD.0 QRegText
PSHD.3
POPD.1
DPUP.1 0d2
CALL breakByte
SETD.0 SRegText
PSHD.3
POPD.1
DPUP.1 0d1
CALL breakByte
CALL newLine
SETD.0 DP0Text
PSHD.3
POPD.1
DPUP.1 0d11
CALL breakWord
SETD.0 DP1Text
PSHD.3
POPD.1
DPUP.1 0d9
CALL breakWord
SETD.0 DP2Text
PSHD.3
POPD.1
DPUP.1 0d7
CALL breakWord
SETD.0 DP3Text
PSHD.3
POPD.1
DPUP.1 0d5
CALL breakWord
CALL newLine
; Anything typed carries on. Reading the data port waits however the console is set, which
; is one key if the program asked for key mode and a whole line if it did not - and either
; way it is the program's own console being borrowed for a moment.
SETD.0 ResumeText
CALL printString
INA 0x00
CALL newLine
RETI
; DP0 names a field and DP1 points at it in the frame. The caller does the stepping, with
; DPUP and a number written into the program, because a routine cannot hand a pointer back:
; CALL saves DP0 to DP2 and RET puts them back, so a walk done in here would be undone on
; the way out. Written that way first, and every field showed the frame's first byte.
breakByte:
CALL printString
LDA.1
CALL printByteHex
INIA 0x20
OUTA 0x00
RET
; The same for the two byte fields, most significant first the way the frame holds them.
breakWord:
CALL printString
LDA.1
CALL printByteHex
INCD.1
LDA.1
CALL printByteHex
INIA 0x20
OUTA 0x00
RET
; A and B together are a number. Prints it in decimal without leading zeroes, which covers
; a line number and a byte count both, so there is no need for one service each.
handlePrintNumber:
SETD.0 PrintNumber
STA.0
INCD.0
STB.0
SETD.0 PrintNumber
CALL printWordDecimal
RETI
; Giving the machine back. This is the one place MVDS earns its keep. The program's Stack,
; and the frame this very interrupt arrived on, are both abandoned where they lie, because
; nothing is going to return through either of them.
;
; Which is exactly why this cannot RETI. Its return address is on the Stack it just walked
; away from, so it branches to the prompt instead.
handleExit:
SETD.1 SystemStack
LDD.0.1
MVDS.0
; Whatever the program put in the vector table comes out again. A vector points into the
; program that supplied it, and the program is gone, so anything left installed would aim
; an interrupt at whatever those addresses hold next.
CALL removeVectors
; The console goes back to how the shell wants it, whatever the program left it in: line
; mode, and not interrupting. A program that wanted either is expected to put it back
; itself, but one that stopped early, or forgot, would otherwise hand back a shell with
; no echo and no backspace, or one being interrupted about keys it is reading anyway.
; Zero is both bits, so this undoes everything the control port can be asked for, and
; asking for what is already the case costs a byte out of a port and does nothing. That
; is the right price for not having to know.
RSTA
OUTA 0x02
SETD.0 Finished
CALL printString
CALL newLine
BRI prompt
; ---- dump ----
;
; dump Sixty four more bytes, carrying on from the last one.
; dump <where> From the start of that bank.
; dump <where> <addr> From there.
;
; <where> is program, data, or a bank number in hexadecimal. That the CPU cannot read
; Program Memory and this can is the whole point: the instruction set has no way to look
; at itself, and the controller does, so a monitor is possible at all only through it.
; b <program|data|number>
;
; The two banks that always exist have names, because "program" is what somebody means and
; 0 is only what the machine calls it. Anything else is a number, and has to be one that is
; really there.
doBank:
SETD.1 TextRest
LDD.0.1
LDA.0
BRA bankWhat
SETD.1 ProgramWord
CALL textSame
BRQ bankProgram
SETD.1 DataWord
CALL textSame
BRQ bankData
CALL textHexWord
BNQ bankWhat
SETD.0 TextValue
INCD.0
LDA.0
BRI bankSet
bankProgram:
RSTA
BRI bankSet
bankData:
INIA 0d1
bankSet:
; The old one is kept, because asking about a bank means writing it down first - and if
; it turns out not to exist, being left pointed at it would fault on the very next look.
SETD.0 DumpBank
LDB.0
SETD.1 BankWas
STB.1
SETD.0 DumpBank
STA.0
; Naming a bank puts the cursor at the start of it, which is the only answer that does
; not depend on what was asked for last time.
CALL bankPresent
BRQ bankNotThere
RSTA
SETD.0 DumpAt
STA.0
INCD.0
STA.0
SETD.0 BankIs
CALL printString
SETD.0 DumpBank
LDA.0
CALL printByteHex
CALL newLine
BRI prompt
bankNotThere:
SETD.0 BankWas
LDA.0
SETD.1 DumpBank
STA.1 ; Back where it was, which is somewhere that exists.
SETD.0 NoSuchBank
CALL printString
CALL newLine
BRI prompt
bankWhat:
SETD.0 BankUsage
CALL printString
CALL newLine
BRI prompt
; x [address] - sixty four bytes. d [address] - eight instructions. Without an address
; either carries on from where the last one stopped, so reading through memory is one
; letter at a time and the two share a place in it.
doExamine:
RSTA
SETD.0 ShowAsCode
STA.0
BRI showAt
doDisassemble:
INIA 0x01
SETD.0 ShowAsCode
STA.0
showAt:
SETD.1 TextRest
LDD.0.1
LDA.0
BRA dumpGo ; Nothing said, so carry on from where the last one stopped.
CALL textHexWord
BNQ dumpBadWhere
SETD.0 TextValue
LDA.0
SETD.1 DumpAt
STA.1
SETD.0 TextValue
INCD.0
LDA.0
SETD.1 DumpAt
INCD.1
STA.1
dumpCheckBank:
CALL bankPresent
BRQ dumpNoBank
dumpGo:
SETD.0 ShowAsCode
LDA.0
BNA disassembleGo
INIA 0d4
SETD.0 DumpRows
STA.0
dumpRow:
SETD.0 DumpAt
CALL printWordHex
INIA 0d2
CALL printSpaces
; Point the controller at the row. Reading the Data port takes a byte and steps the
; source on, so the whole row is one instruction repeated.
SETD.0 DumpBank
LDA.0
OUTA 0xE0
SETD.0 DumpAt
LDA.0
OUTA 0xE1
INCD.0
LDA.0
OUTA 0xE2
; Sixteen bytes, kept as they go past so that they can be shown twice.
INIA 0d16
SETD.0 DumpCount
STA.0
SETD.1 DumpBytes
dumpByte:
INA 0xE9
STA.1
CALL printByteHex
INIA 0x20
OUTA 0x00
INCD.1
SETD.0 DumpCount
LDA.0
DECA
STA.0
BNA dumpByte
; The same sixteen again, as characters. Anything that is not printable shows as a dot,
; because a control character sent to the console would move the cursor and ruin the
; shape of the dump.
INIA 0x20
OUTA 0x00
INIA 0d16
SETD.0 DumpCount
STA.0
SETD.1 DumpBytes
dumpChar:
LDA.1
INIB 0x20
CCF
SUB
BRC dumpDot ; Below a space.
INIB 0x7F
CCF
SUB
BNC dumpDot ; Delete, or above it.
OUTA 0x00
BRI dumpCharNext
dumpDot:
INIA 0x2E
OUTA 0x00
dumpCharNext:
INCD.1
SETD.0 DumpCount
LDA.0
DECA
STA.0
BNA dumpChar
CALL newLine
; Sixteen further along, carrying into the high byte if the low one wrapped.
SETD.0 DumpAt
INCD.0
LDA.0
INIB 0d16
CCF
ADD
STQ.0
BNC dumpRowNext
SETD.0 DumpAt
LDA.0
INCA
STA.0
dumpRowNext:
SETD.0 DumpRows
LDA.0
DECA
STA.0
BNA dumpRow
BRI prompt
disassembleGo:
INIA 0d8
SETD.0 DumpRows
STA.0
disassembleOne:
CALL showInstruction
SETD.0 DumpRows
LDA.0
DECA
STA.0
BNA disassembleOne
BRI prompt
dumpBadWhere:
SETD.0 ExamineUsage
CALL printString
CALL newLine
BRI prompt
dumpNoBank:
SETD.0 NoSuchBank
CALL printString
CALL newLine
BRI prompt
; s <address> <byte> <byte> ...
;
; Writes into whichever bank is being looked at, THROUGH THE CONTROLLER, so Program Memory
; can be changed as easily as Data - which no instruction on this machine can do, and which
; is most of the reason for having a monitor at all.
;
; The cursor is left alone. Somebody poking a byte is usually looking at something else, and
; having the address they were reading move underneath them would be a poor reward.
doSet:
; A bank can be present and still refuse to be written: the controller's own table is
; published read only, and writing to it is refused. A refusal nobody catches stops the
; machine, which is a poor answer to somebody looking around with b and then typing s.
CALL bankPresent
SETD.0 BankFlags
LDA.0
INIB 0x02 ; The read only bit of that bank's record.
AND
BNQ setReadOnly
SETD.1 TextRest
LDD.0.1
CALL textHexWord
BNQ setWhat
SETD.1 DumpBank
LDA.1
OUTA 0xE3
SETD.1 TextValue
LDA.1
OUTA 0xE4
INCD.1
LDA.1
OUTA 0xE5
CALL stepPastNumber
PSHD.3
POPD.0
setByte:
CALL textHexWord
BNQ prompt ; Nothing more on the line, so that was all of them.
SETD.1 TextValue
INCD.1
LDA.1
OUTA 0xE9 ; The destination steps on by itself, so a run of bytes is a loop.
CALL stepPastNumber
PSHD.3
POPD.0
BRI setByte
setReadOnly:
SETD.0 ReadOnlyText
CALL printString
CALL newLine
BRI prompt
setWhat:
SETD.0 SetUsage
CALL printString
CALL newLine
BRI prompt
; g <address>
;
; Somewhere to go. It does not come back by itself - that would want a breakpoint, which is
; a byte written over an instruction and a handler waiting for it, and neither exists yet.
; But a program that gives the machine back the ordinary way lands at the prompt it was
; started from, which is this one, still in the monitor.
doGo:
SETD.1 TextRest
LDD.0.1
CALL textHexWord
BNQ goWhat
; Where the Stack is now, written down before leaving, exactly as run does it. Whatever is
; jumped to may give the machine back through osExit, and osExit puts the Stack back to
; what is written here - so without this it would restore the one the LAST run left, or
; none at all, and the shell would come back with its Stack pointing at nothing.
MVSD.0
SETD.1 SystemStack
STD.0.1
SETD.1 TextValue
LDD.3.1
BRD.3
goWhat:
SETD.0 GoUsage
CALL printString
CALL newLine
BRI prompt
; DP0 names text that textHexWord has just read a number off the front of. LEAVES DP3 past
; the digits and any spaces after them, ready for the next one.
;
; DP3 rather than DP0, because a subroutine cannot hand a pointer back in DP0: CALL saves it
; and RET puts it back, so stepping it here would be undone on the way out.
stepPastNumber:
PSHD.0
POPD.3
SETD.1 TextDigits
LDB.1
stepPastDigits:
BRB stepPastSpaces
INCD.3
DECB
BRI stepPastDigits
stepPastSpaces:
LDA.3
BRA stepPastDone
INIB 0x20
XOR
BNQ stepPastDone
INCD.3
BRI stepPastSpaces
stepPastDone:
RET
; Points the controller's source at the cursor, so that reading port 0xE9 walks forwards.
aimAtDumpAt:
SETD.0 DumpBank
LDA.0
OUTA 0xE0
SETD.0 DumpAt
LDA.0
OUTA 0xE1
INCD.0
LDA.0
OUTA 0xE2
RET
; One byte from the cursor, and the cursor moves on.
;
; THE BYTE COMES BACK IN Q, not in A, because a subroutine cannot hand anything back in A:
; CALL saves it and RET puts it back, so an assignment here would be undone by the return.
takeByte:
INA 0xE9
PSHA
SETD.0 DumpAt
INCD.0
LDA.0
INCA
STA.0
BNC takeByteDone
DECD.0
LDA.0
INCA
STA.0
takeByteDone:
POPA
RSTB
OR
RET
; ---- Showing instructions ----
;
; The half of a monitor that a byte dump cannot do. What it needs and a dump does not
; is to know how LONG each instruction is, because getting that wrong does not print one
; line wrong - it loses the place and prints everything after it wrong.
; One instruction: where it is, the bytes it is made of, and what it says.
showInstruction:
SETD.0 DumpAt
LDA.0
CALL printByteHex
INCD.0
LDA.0
CALL printByteHex
INIA 0x20
OUTA 0x00
INIA 0x20
OUTA 0x00
CALL aimAtDumpAt
CALL takeByte
MVQA
SETD.0 Opcode
STA.0
CALL findInstruction
BNQ showUnknown
; What shape it is, and from that how many bytes it runs to.
PSHD.3
POPD.0
INCD.0
LDA.0
SETD.1 Shape
STA.1
SETD.0 ShapeLength
LDB.1
shapeStep:
BRB shapeGot
INCD.0
DECB
BRI shapeStep
shapeGot:
LDA.0
SETD.1 Length
STA.1
; The rest of its bytes. The first one is already read.
SETD.0 InstrBytes
SETD.1 Opcode
LDA.1
STA.0
INCD.0
SETD.1 Length
LDB.1
DECB
readRest:
BRB readRestDone
PSHB
CALL takeByte
POPB
MVQA
STA.0
INCD.0
DECB
BRI readRest
readRestDone:
; Show them, padded out so that what follows lines up however long the instruction was.
SETD.0 InstrBytes
SETD.1 Length
LDB.1
showBytes:
PSHB
LDA.0
CALL printByteHex
INIA 0x20
OUTA 0x00
POPB
INCD.0
DECB
BNB showBytes
INIB 0d4
SETD.0 Length
LDA.0
padBytes:
CCF
SUB
BRQ padDone ; As many as there are, so nothing to pad.
PSHA
PSHB
INIA 0x20
OUTA 0x00
OUTA 0x00
OUTA 0x00
POPB
POPA
DECB
BRI padBytes
padDone:
INIA 0x20
OUTA 0x00
; Its name, without the spaces it is padded to four with.
PSHD.3
POPD.0
DPUP.0 0d02
INIB 0d4
showName:
LDA.0
BRA showNameDone
PSHB
INIB 0x20
XOR
POPB
BRQ showNameDone
OUTA 0x00
INCD.0
DECB
BNB showName
showNameDone:
; And whatever follows it, which depends only on the shape.
SETD.0 Shape
LDA.0
BRA showOperandNone ; 0, nothing at all
DECA
BRA showAddress ; 1, an address
DECA
BRA showByte ; 2, one byte
DECA
BRA showSelector ; 3, a Data Pointer
DECA
BRA showSelectorByte ; 4, a Data Pointer and a byte
DECA
BRA showSelectorAddress ; 5, a Data Pointer and an address
BRI showTwoSelectors ; 6
showOperandNone:
CALL newLine
RET
showAddress:
INIA 0x20
OUTA 0x00
SETD.0 InstrBytes
INCD.0
LDA.0
CALL printByteHex
INCD.0
LDA.0
CALL printByteHex
CALL newLine
RET
showByte:
INIA 0x20
OUTA 0x00
SETD.0 InstrBytes
INCD.0
LDA.0
CALL printByteHex
CALL newLine
RET
showSelector:
CALL putSelectorOne
CALL newLine
RET
showSelectorByte:
CALL putSelectorOne
INIA 0x20
OUTA 0x00
SETD.0 InstrBytes
DPUP.0 0d02
LDA.0
CALL printByteHex
CALL newLine
RET
showSelectorAddress:
CALL putSelectorOne
INIA 0x20
OUTA 0x00
SETD.0 InstrBytes
DPUP.0 0d02
LDA.0
CALL printByteHex
INCD.0
LDA.0
CALL printByteHex
CALL newLine
RET
showTwoSelectors:
CALL putSelectorOne
INIA 0d46 ; .
OUTA 0x00
SETD.0 InstrBytes
DPUP.0 0d02
LDA.0
CALL printDecimalDigit
CALL newLine
RET
; ".n" for the selector that follows the opcode.
putSelectorOne:
INIA 0d46
OUTA 0x00
SETD.0 InstrBytes
INCD.0
LDA.0
CALL printDecimalDigit
RET
; A byte that decodes as nothing. Shown as it is, and the cursor moves on by one, because
; the only honest thing to do with a byte that is not an instruction is say so and carry on.
showUnknown:
SETD.0 Opcode
LDA.0
CALL printByteHex
SETD.0 UnknownText
SWI osPrintString
RET
; Looks the opcode up. DP3 lands on its entry and Q is zero, or Q is not zero and it is not
; an instruction at all.
findInstruction:
SETD.3 Instructions
SETD.0 InstructionCount
LDB.0
findStep:
LDA.3
SETD.0 Opcode
PSHB
LDB.0
XOR
POPB
BRQ findFound
DPUP.3 0d07
DECB
BNB findStep
RSTA
INIB 0d1
CCF
ADD
RET
findFound:
RSTA
RSTB
CCF
ADD
RET
; Is there such a bank? Q is zero if there is not.
;
; ASKING THE CONTROLLER FOR A BANK THAT IS NOT THERE IS REFUSED, and a refusal nobody
; catches stops the machine, which is a wretched answer to a mistyped number. The bank table
; says what exists, and it lives in bank 2.
;
; Bank n's record starts at n times eight. A and B are a shift register sixteen bits wide,
; so putting the number in the low half and rotating left three times multiplies it by
; eight without anything falling off the top: the most it can reach is 2040.
bankPresent:
RSTA
SETD.0 DumpBank
LDB.0
SHL SHL SHL
SETD.0 DumpRecord
STA.0
INCD.0
STB.0
INIA 0d2
OUTA 0xE0 ; SourceBank: the controller's own memory.
SETD.0 DumpRecord
LDA.0
OUTA 0xE1
INCD.0
LDA.0
OUTA 0xE2
INA 0xE9 ; The flags byte of that bank's record.
SETD.0 BankFlags
STA.0 ; Kept whole, since present is not the only thing it says.
INIB 0x01
AND
RET
; ---- help ----
doHelp:
SETD.0 HelpText
CALL printString
CALL newLine
SETD.0 HelpMoreText
CALL printString
CALL newLine
; And what the monitor adds, but only when it is on. Listing commands that would not
; answer is a way of teaching somebody something untrue.
SETD.0 Mode
LDA.0
BRA prompt
SETD.0 MonitorHelp
CALL printString
CALL newLine
BRI prompt
#Data
Banner:
"CosmOS"
PromptText:
"> "
NoDisk:
"no filesystem on the disk"
Unknown:
"I do not know: "
Farewell:
"halted"
FilesText:
" files"
FileText:
" file"
; Two strings rather than one, because a string literal stops at 255 characters and each
; one carries its own zero byte, so they are printed in turn rather than joined.
HelpText:
"dir list what is on the disk
load <file> read a program off the disk
run [words] start what was loaded, and tell it those words
delete <file> take it off the disk
rename <file> <to> call it something else"
HelpMoreText:
"monitor look at memory, change it, and jump into it
help this
exit stop, or leave the monitor if you are in it"
ExamineUsage:
"x <address>, or x on its own to carry on"
NoSuchBank:
"there is no such bank"
ProgramWord:
"program"
DataWord:
"data"
ExecMagic:
"SBEX"
LoadWhat:
"load what?"
NoSuchFile:
"no such file"
Deleted:
"gone"
Renamed:
"renamed"
DeleteWhat:
"delete what?"
RenameWhat:
"rename what to what?"
RenameNo:
"there is no such file, or that name is taken"
TooManyVectors:
"that program wants more vectors than there is room for"
Unreadable:
"could not read it"
NotProgram:
"not a program"
WrongVersion:
"a version I do not know"
LoadedText:
"loaded, starting at "
NothingLoaded:
"nothing is loaded"
Finished:
"finished"
BreakText:
"break at "
ARegText:
"A "
BRegText:
"B "
QRegText:
"Q "
SRegText:
"S "
DP0Text:
"DP0 "
DP1Text:
"DP1 "
DP2Text:
"DP2 "
DP3Text:
"DP3 "
ResumeText:
"press a key "
MonitorPrompt:
"* "
UnknownText:
" ?
"
MonitorName:
"monitor"
MonitorHelp:
"x examine, d disassemble, s set, b bank, g go, exit leaves"
ExamineName:
"x"
DisName:
"d"
SetName:
"s"
BankName:
"b"
GoName:
"g"
BankUsage:
"b <program|data|number>"
BankIs:
"bank "
SetUsage:
"s <address> <byte> <byte> ..."
ReadOnlyText:
"that bank will not be written"
GoUsage:
"g <address>"
DirName:
"dir"
LoadName:
"load"
RunName:
"run"
DeleteName:
"delete"
RenameName:
"rename"
HelpName:
"help"
ExitName:
"exit"
DiskReady:
0x00
LoadedOk:
0x00
LoadedEntry:
0x00 0x00
LoadCount:
0x00
LoadVersion:
0x00
; Where the first of rename's two names is, kept while the second is picked out of the
; line, since finding that needs the pointers for itself.
RenameFrom:
0x00 0x00
; What followed "run", kept for the program to ask for.
RunArgument:
#Reserve 0d64
; A number on its way to being printed, since the routine that prints one wants it in
; memory and a service is handed it in registers.
PrintNumber:
0x00 0x00
; ---- The vectors a loaded program brought with it ----
;
; Six bytes each: where it goes, what goes there, and what was there before. The last two
; are filled in when the program runs and read back when it exits, so what a program covers
; up comes back rather than becoming a hole.
;
; Sixteen is a limit rather than a considered number. It is far more than anything written
; so far wants, and a program asking for more is refused at load rather than having some of
; its handlers installed and the rest dropped.
VectorSource:
0x00 0x00
VectorsLeft:
0x00
LoadedVectorCount:
0x00
LoadedVectors:
#Reserve 0d96
; Where the monitor is looking, so that a bare 'dump' can carry on from it.
DumpBank:
0x00
DumpAt:
0x00 0x00
DumpRows:
0x00
Mode:
0x00
DumpCount:
0x00
BankWas:
0x00
BankFlags:
0x00
ShowAsCode:
0x00
DumpRecord:
0x00 0x00
Opcode:
0x00
Shape:
0x00
Length:
0x00
InstrBytes:
#Reserve 0d4
; How many bytes an instruction of each shape runs to, the opcode included.
ShapeLength:
0d1 0d3 0d2 0d2 0d3 0d4 0d3
InstructionCount:
0d64
; ---- The instruction table ----
;
; Generated by Tests/instructiontable.py from the assembler's own list, and checked against
; it by Tests/docs.sh. Seven bytes each: the opcode, the shape, and four characters of name
; with the zero the assembler puts after a string.
Instructions:
0x00 0d0 "ADD "
0x01 0d0 "SUB "
0x02 0d0 "AND "
0x03 0d0 "OR "
0x04 0d0 "XOR "
0x05 0d0 "NOTA"
0x06 0d0 "NOTB"
0x07 0d0 "SHL "
0x08 0d0 "SHR "
0x10 0d1 "BRI "
0x11 0d1 "BRQ "
0x12 0d1 "BRA "
0x13 0d1 "BRB "
0x14 0d1 "BRC "
0x15 0d3 "BRD "
0x1A 0d1 "BNQ "
0x1B 0d1 "BNA "
0x1C 0d1 "BNB "
0x1D 0d1 "BNC "
0x17 0d1 "CALL"
0x18 0d2 "SWI "
0x19 0d0 "RETI"
0x1F 0d0 "RET "
0x20 0d0 "RSTA"
0x21 0d0 "RSTB"
0x22 0d0 "INCA"
0x23 0d0 "INCB"
0x24 0d0 "DECA"
0x25 0d0 "DECB"
0x26 0d2 "INIA"
0x27 0d2 "INIB"
0x28 0d0 "CCF "
0x29 0d0 "MVQA"
0x2A 0d0 "MVQB"
0x2B 0d0 "SIF "
0x2C 0d0 "CIF "
0x30 0d0 "PSHQ"
0x31 0d0 "PSHA"
0x32 0d0 "PSHB"
0x33 0d3 "PSHD"
0x34 0d0 "POPA"
0x35 0d0 "POPB"
0x36 0d3 "POPD"
0x40 0d3 "INCD"
0x41 0d3 "DECD"
0x42 0d3 "LDA "
0x43 0d3 "LDB "
0x44 0d3 "STQ "
0x45 0d3 "STA "
0x46 0d3 "STB "
0x47 0d5 "SETD"
0x48 0d4 "DPUP"
0x49 0d4 "DPDN"
0x4A 0d6 "LDD "
0x4B 0d6 "STD "
0x4C 0d3 "MVSD"
0x4D 0d3 "MVDS"
0xD0 0d2 "OUTQ"
0xD1 0d2 "OUTA"
0xD2 0d2 "OUTB"
0xE0 0d2 "INA "
0xE1 0d2 "INB "
0xF0 0d0 "NOP "
0xFF 0d0 "HALT"
DumpBytes:
#Reserve 0d16
; Where the system's Stack was when it handed the machine to a program. Kept below the
; region a program owns, so that a program has to go looking to break it.
SystemStack:
0x00 0x00
DirSeen:
0x00
DirSize:
0x00 0x00
WidthLeft:
0x00
; Sixty three characters and the zero byte that ends them.
CommandLine:
#Reserve 0d64
#Vectors
Boot boot
osPrintString handlePrintString
osReadLine handleReadLine
osExit handleExit
osArgument handleArgument
osFileRead handleFileRead
osFileSave handleFileSave
osFileDelete handleFileDelete
osFileRename handleFileRename
osPrintNumber handlePrintNumber
osBreak handleBreak
Device 0x20 diskDone
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