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SplitBit-Emulator/Programs/CosmOS/Apps/Stream.asm
T
Anachronaut 87d819847e A program can say how it went 
SWI osExit takes a status in A, and the shell keeps it. Fifty eight exits
across twenty three programs now say deliberately whether they worked: 25
did what they were asked, 24 did not, 9 were asked wrongly. Compare is the
exception and says so - one there means the files differ, which is a result
rather than a failure, the way diff has always had it.

IN A RATHER THAN Q, which is not a departure from the rule that a service
answers in Q. This one takes an ARGUMENT, the way osPrintNumber takes A and
B, and it never returns to answer anything. A is free precisely because a
return would have put it back - and Q is the ALU's output, so a small
number costs four instructions there against one in A.

The shell does not print it. A program that failed has already said so in
words and a number beside that is noise, so osLastStatus hands it back and
Status is the program that shows it. That indirection is the point: the
number exists for the thing that cannot read words.

MARKING THE EXITS FOUND A DEFECT ON THE FIRST RUN. Type and More printed
why they had failed and then fell through into the success exit, reporting
that all was well. Nobody had noticed, because while the only reader was a
person, the person could see both the complaint and the claim.

Two smaller things. Snake sets the console to line mode and then exits with
zero, and the linter flagged the second RSTA as redundant - an exit status
and a console mode, equal by accident, which is the class that must never
be collapsed. And the README still taught answering by writing into the
frame, three months of habit that SRET replaced yesterday; that section is
gone and the one describing SRET stands in its place.
2026-08-27 19:16:21 -04:00

496 lines
10 KiB
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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 0x4000
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 0x2000
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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