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SplitBit-Emulator/Programs/CosmOS/Apps/Stream.asm
T
AnachronautandClaude Opus 5 3449405b18 Give the system another page of each memory 
CosmOS had 1,161 bytes of Program Memory left before the address applications
load at, and Tab completion is not going to fit in that with anything to spare.
So the wall moves up one page: the system keeps below 0x4FFF and 0x2FFF, and an
application is based at 0x5000 and 0x3000.

A PAGE IS A CHEAP THING TO GIVE IT AND AN EXPENSIVE THING TO RUN OUT OF. An
application still has 44K of Program Memory before the vector table and the
largest one here uses 7.5K, so what was taken from applications is space nothing
has ever asked for - while what the system gained is the difference between
building the next thing and counting bytes while building it.

Not doubling, which was the version that would have cost application space worth
minding. One page, and the same again when it is needed.

Nothing in the machine knows where the wall is, so this is 34 #Base lines, one
threshold in the fault handler, and the table in the CosmOS README that
Tests/docs.sh reads its limits out of.

The native assembler's scratch map had to move with it, and docs.sh said so
before anything ran: its data reached 0x40D6 and its buffers began at 0x4000, so
they were sitting on its variables. That file already carries a paragraph about
the floor coming up and the map staying where it was. It has happened twice now,
and been caught by a check the first time wrote.

Twenty three recordings are the same runs a page higher.

Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_01E2JrLzFvuFX9fgi1LDRjrW
2026-09-01 16:49:12 -04:00

496 lines
10 KiB
NASM
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; Reading a file the machine cannot hold.
;
; Every other program here asks for a file and is handed the whole of it, which settles the
; question for anything under 64K and settles nothing above. CosmOS's own source is above:
; the sources together are a hundred kilobytes, and Data Memory is sixty four. A machine
; that is one day going to assemble itself has to be able to read a file bigger than its
; memory, and this is the program that proves it can.
;
; It uses osFileInfo and osFileBlock, and nothing else knows how a filesystem works. There
; is no open and no close - every call names the file and says which block it wants, so a
; program that stops halfway leaves nothing behind for anybody to clean up.
;
; ---- What it checks, and why each one is here ----
;
; 1. A file of four hundred odd blocks is read from end to end, a block at a time, into a
; buffer of one block. That is the feature.
; 2. A small file is read BOTH WAYS - whole with osFileRead, and streamed - and the two
; have to agree. This is the real proof: it compares streaming against the path that
; was already known to work, so a fault in the block count or the order of the blocks
; shows up as a difference rather than as a plausible wrong answer.
; 3. Two files are read alternately. The system remembers where the last file it was
; asked about lives, and this is the case that catches a memory that does not notice
; the name has changed.
; 4. A rename in the middle. Same reason, from the other side: the file the system
; remembers has moved out from under the name it remembered it by.
; 5. The three ways of being told no, each with its own number.
;
; THE CHECKSUM IS FLETCHER'S, not a sum. A plain total is the same whatever order the bytes
; arrived in, and the order is exactly what streaming has to get right; carrying a second
; accumulator that adds the first one in each time makes a block delivered out of turn
; change the answer.
;
; Written by Anachronaut
#Include services.asm
#Program
#Base 0x5000
start:
; ---- 1. How big is something that will not fit ----
;
; In blocks, not bytes, and that is forced rather than chosen: a file on a sixteen
; megabyte disk can be twenty four bits long and a pointer holds sixteen.
SETD.0 BigName
SWI osFileInfo
BNQ noBig
SETD.0 BigIs
SWI osPrintString
PSHD.3
POPB
POPA
SWI osPrintNumber
SETD.0 BlocksText
SWI osPrintString
; ---- 2. Read the whole of it through a hole one block wide ----
CALL clearChecksum
CALL clearIndex
bigLoop:
SETD.0 BigName
SETD.1 Block
SETD.2 Index
LDA.2
INCD.2
LDB.2 ; Which block, most significant first.
SWI osFileBlock
BNQ bigDone
CALL takeCount
SETD.1 Block
CALL checksum
CALL stepIndex
BRI bigLoop
bigDone:
; The loop ends because a block past the end was asked for, which is answer three. Any
; other answer stopped it early and would otherwise look exactly like success, so what
; ended it is printed rather than assumed.
CALL keepWhy
SETD.0 ReadText
SWI osPrintString
SETD.2 Index
LDA.2
INCD.2
LDB.2
SWI osPrintNumber
SETD.0 BlocksSumText
SWI osPrintString
CALL printChecksum
SETD.0 StoppedText
SWI osPrintString
CALL printWhy
; ---- 3. The same file both ways ----
;
; osFileRead is the path that already worked, so it is what streaming is measured
; against. If the two checksums agree, every byte arrived and they arrived in order.
SETD.0 SmallName
SETD.1 Whole
SWI osFileRead
BNQ noSmall
CALL takeCount
CALL clearChecksum
SETD.1 Whole
CALL checksum
CALL keepChecksum
CALL clearChecksum
CALL clearIndex
smallLoop:
SETD.0 SmallName
SETD.1 Block
SETD.2 Index
LDA.2
INCD.2
LDB.2
SWI osFileBlock
BNQ smallDone
CALL takeCount
SETD.1 Block
CALL checksum
CALL stepIndex
BRI smallLoop
smallDone:
SETD.0 BothText
SWI osPrintString
CALL printChecksum
SETD.0 AgainstText
SWI osPrintString
CALL printKept
SETD.0 NewLine
SWI osPrintString
CALL sameAsKept
BNQ differ
SETD.0 SameText
SWI osPrintString
BRI interleave
differ:
SETD.0 DifferText
SWI osPrintString
; ---- 4. Two files, alternately ----
;
; Block zero of the big file, then a block of the small one, then block zero of the big
; file again. The two readings of the same block have to match. A system that remembered
; the first file and did not notice the name had changed would hand back a block of the
; wrong file in the middle, and then the right one again, so only the middle call would
; be wrong - which is why this asks for the same block twice rather than once.
interleave:
CALL clearChecksum
CALL readFirstBig
CALL keepChecksum
SETD.0 SmallName
SETD.1 Block
RSTA
RSTB
SWI osFileBlock
CALL clearChecksum
CALL readFirstBig
CALL sameAsKept
BNQ mixedUp
SETD.0 InterleaveOk
SWI osPrintString
BRI moved
mixedUp:
SETD.0 InterleaveBad
SWI osPrintString
; ---- 5. A file that moves out from under the name ----
;
; The system has just been asked about the small file, so it is the one being remembered.
; Renaming it has to throw that away: the blocks are still there and still hold the same
; bytes, so a stale answer would work perfectly and be wrong.
moved:
SETD.0 SmallName
SETD.1 OtherName
SWI osFileRename
BNQ noRename
SETD.0 MovedText
SWI osPrintString
SETD.0 SmallName
SWI osFileInfo
CALL keepWhy
SETD.0 OldNameText
SWI osPrintString
CALL printWhy
SETD.0 NewNameText
SWI osPrintString
SETD.0 OtherName
SWI osFileInfo
CALL keepWhy
CALL printWhy
; ---- 6. The three ways of being told no ----
missing:
SETD.0 MissingName
SWI osFileInfo
CALL keepWhy
SETD.0 MissingText
SWI osPrintString
CALL printWhy
SETD.0 OtherName
SETD.1 Block
INIA 0xFF
INIB 0xFF
SWI osFileBlock
CALL keepWhy
SETD.0 PastText
SWI osPrintString
CALL printWhy
INIA 0d1
SWI osExit
noBig:
CALL keepWhy
SETD.0 NoBigText
SWI osPrintString
CALL printWhy
INIA 0d1
SWI osExit
noSmall:
SETD.0 NoSmallText
SWI osPrintString
INIA 0d1
SWI osExit
noRename:
SETD.0 NoRenameText
SWI osPrintString
INIA 0d1
SWI osExit
; ---- Routines ----
; Block zero of the big file, into the running checksum.
readFirstBig:
SETD.0 BigName
SETD.1 Block
RSTA
RSTB
SWI osFileBlock
BNQ readFirstDone
CALL takeCount
SETD.1 Block
CALL checksum
readFirstDone:
RET
; What the service just answered in DP3 becomes Left, which is what the checksum counts
; down. Kept in memory rather than in a pointer because a CALL does not preserve one.
takeCount:
PSHD.3
POPB
POPA
SETD.2 Left
STA.2
INCD.2
STB.2
RET
; Adds the bytes at DP1 into the running checksum, as many of them as Left says.
;
; Two accumulators, each a byte wide, each throwing away what carries off the top. The
; first is the sum of the bytes and the second is the sum of the first, so a byte that
; arrives late counts for less than one that arrived early - which is what makes this
; notice a block delivered out of turn.
checksum:
checksumLoop:
LDA.1
SETD.2 Fletch1
LDB.2
CCF
ADD
MVQA
STA.2
SETD.2 Fletch2
LDB.2
CCF
ADD
MVQA
STA.2
INCD.1
; Left goes down by one, sixteen bits of it: a whole block is 256 bytes and a whole file
; is more than one block, so a byte counter would not reach.
SETD.2 Left
INCD.2
LDA.2
BNA checksumLow
DECD.2
LDA.2
DECA
STA.2 ; Borrow out of the high byte.
INCD.2
INIA 0xFF
STA.2
BRI checksumTest
checksumLow:
DECA
STA.2
checksumTest:
SETD.2 Left
LDA.2
INCD.2
LDB.2
OR ; Zero only when both halves are.
BNQ checksumLoop
RET
clearChecksum:
RSTA
SETD.2 Fletch1
STA.2
SETD.2 Fletch2
STA.2
RET
clearIndex:
RSTA
SETD.2 Index
STA.2
INCD.2
STA.2
RET
stepIndex:
SETD.2 Index
INCD.2
LDA.2
INCA
STA.2
BNC stepIndexDone
DECD.2
LDA.2
INCA
STA.2
stepIndexDone:
RET
; Puts the checksum aside so that a second one can be compared with it.
keepChecksum:
SETD.2 Fletch1
LDA.2
SETD.2 Kept1
STA.2
SETD.2 Fletch2
LDA.2
SETD.2 Kept2
STA.2
RET
; Q is zero if the running checksum is the one that was put aside.
sameAsKept:
SETD.2 Fletch1
LDA.2
SETD.2 Kept1
LDB.2
XOR
BNQ sameAsKeptDone
SETD.2 Fletch2
LDA.2
SETD.2 Kept2
LDB.2
XOR
sameAsKeptDone:
RET
printChecksum:
SETD.2 Fletch1
LDA.2
SETD.2 Fletch2
LDB.2
SWI osPrintNumber
RET
printKept:
SETD.2 Kept1
LDA.2
SETD.2 Kept2
LDB.2
SWI osPrintNumber
RET
; Why the last service said no. Q survives a CALL, which is the only reason this can be a
; routine at all, but it does not survive the next SWI - so it is written down here and
; printed later, with whatever has to happen in between happening in between.
keepWhy:
MVQA
SETD.2 Why
STA.2
RET
printWhy:
RSTA
SETD.2 Why
LDB.2
SWI osPrintNumber
SETD.0 NewLine
SWI osPrintString
RET
#Data
#Base 0x3000
BigName:
"big.txt"
SmallName:
"small.txt"
OtherName:
"moved.txt"
MissingName:
"nothing.txt"
BigIs:
"big.txt is "
BlocksText:
" blocks
"
ReadText:
"read "
BlocksSumText:
" blocks, checksum "
StoppedText:
", stopped with "
BothText:
"small.txt streamed is "
AgainstText:
", read whole is "
SameText:
"the same
"
DifferText:
"DIFFERENT
"
InterleaveOk:
"the same block twice with another file between: the same
"
InterleaveBad:
"the same block twice with another file between: DIFFERENT
"
MovedText:
"renamed small.txt
"
OldNameText:
"the old name now answers "
NewNameText:
"the new name answers "
MissingText:
"a name that was never there answers "
PastText:
"a block past the end answers "
NoBigText:
"big.txt would not open, answer "
NoSmallText:
"small.txt would not read
"
NoRenameText:
"it would not rename
"
NewLine:
"
"
Index:
0x00 0x00
Left:
0x00 0x00
Fletch1:
0x00
Fletch2:
0x00
Kept1:
0x00
Kept2:
0x00
Why:
0x00
; One block, which is the whole point: the big file is four hundred times this.
Block:
#Reserve 0d256
; And room for the small one all at once, so that the two ways of reading it can be
; compared against each other.
Whole:
#Reserve 0d1024
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