AnachronautandClaude Opus 5
588e02aff5
The memory map gave CosmOS 0x0000 through 0x1FFF of Program Memory and applications 0x2000 and above. CosmOS is 8141 bytes at the previous commit, which is fifty one bytes short of the line, and the next thing added to it went over. GOING OVER DOES NOT FAIL WHERE IT HAPPENS. Nothing enforces the division: an application says where it goes with #Base and the loader puts it there, so a CosmOS that has grown past 0x1FFF simply has the next program loaded written over the end of it. What breaks is whichever part of the shell that program happened to cover, at whatever later moment somebody uses it. It turned up here as the monitor's assemble command answering "I do not know" to valid instructions, several commands into a session, on a machine that had booted perfectly well. Both halves are doubled: applications now start at 0x4000 in Program Memory and 0x2000 in Data Memory. That is 16K of code and 8K of data for the system, against the 8775 and 2948 it uses today. Both were on the same trajectory, and moving them together means the twenty files that say #Base are edited once rather than twice. The standalone loader's loadable.asm keeps its old base: it belongs to the loader CosmOS grew out of, not to CosmOS, and its addresses answer to a different program. The unbased-segment diagnostic keeps its old base too - it exists to produce an error message that names the address, and the message is what is recorded. Tests/docs.sh now reads the two limits out of the table in the README and measures both segments against them. It reads them rather than being told them because the table is the specification, and this is the second time in this project that the thing nobody checked is the thing that rotted. Co-Authored-By: Claude Opus 5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_01E2JrLzFvuFX9fgi1LDRjrW
492 lines
10 KiB
NASM
492 lines
10 KiB
NASM
; Reading a file the machine cannot hold.
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;
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; Every other program here asks for a file and is handed the whole of it, which settles the
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; question for anything under 64K and settles nothing above. CosmOS's own source is above:
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; the sources together are a hundred kilobytes, and Data Memory is sixty four. A machine
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; that is one day going to assemble itself has to be able to read a file bigger than its
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; memory, and this is the program that proves it can.
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;
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; It uses osFileInfo and osFileBlock, and nothing else knows how a filesystem works. There
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; is no open and no close - every call names the file and says which block it wants, so a
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; program that stops halfway leaves nothing behind for anybody to clean up.
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;
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; ---- What it checks, and why each one is here ----
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;
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; 1. A file of four hundred odd blocks is read from end to end, a block at a time, into a
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; buffer of one block. That is the feature.
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; 2. A small file is read BOTH WAYS - whole with osFileRead, and streamed - and the two
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; have to agree. This is the real proof: it compares streaming against the path that
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; was already known to work, so a fault in the block count or the order of the blocks
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; shows up as a difference rather than as a plausible wrong answer.
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; 3. Two files are read alternately. The system remembers where the last file it was
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; asked about lives, and this is the case that catches a memory that does not notice
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; the name has changed.
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; 4. A rename in the middle. Same reason, from the other side: the file the system
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; remembers has moved out from under the name it remembered it by.
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; 5. The three ways of being told no, each with its own number.
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;
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; THE CHECKSUM IS FLETCHER'S, not a sum. A plain total is the same whatever order the bytes
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; arrived in, and the order is exactly what streaming has to get right; carrying a second
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; accumulator that adds the first one in each time makes a block delivered out of turn
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; change the answer.
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;
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; Written by Anachronaut
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#Include services.asm
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#Program
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#Base 0x4000
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start:
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; ---- 1. How big is something that will not fit ----
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;
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; In blocks, not bytes, and that is forced rather than chosen: a file on a sixteen
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; megabyte disk can be twenty four bits long and a pointer holds sixteen.
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SETD.0 BigName
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SWI osFileInfo
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BNQ noBig
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SETD.0 BigIs
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SWI osPrintString
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PSHD.3
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POPB
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POPA
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SWI osPrintNumber
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SETD.0 BlocksText
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SWI osPrintString
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; ---- 2. Read the whole of it through a hole one block wide ----
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CALL clearChecksum
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CALL clearIndex
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bigLoop:
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SETD.0 BigName
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SETD.1 Block
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SETD.2 Index
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LDA.2
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INCD.2
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LDB.2 ; Which block, most significant first.
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SWI osFileBlock
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BNQ bigDone
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CALL takeCount
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SETD.1 Block
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CALL checksum
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CALL stepIndex
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BRI bigLoop
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bigDone:
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; The loop ends because a block past the end was asked for, which is answer three. Any
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; other answer stopped it early and would otherwise look exactly like success, so what
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; ended it is printed rather than assumed.
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CALL keepWhy
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SETD.0 ReadText
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SWI osPrintString
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SETD.2 Index
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LDA.2
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INCD.2
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LDB.2
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SWI osPrintNumber
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SETD.0 BlocksSumText
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SWI osPrintString
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CALL printChecksum
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SETD.0 StoppedText
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SWI osPrintString
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CALL printWhy
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; ---- 3. The same file both ways ----
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;
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; osFileRead is the path that already worked, so it is what streaming is measured
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; against. If the two checksums agree, every byte arrived and they arrived in order.
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SETD.0 SmallName
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SETD.1 Whole
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SWI osFileRead
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BNQ noSmall
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CALL takeCount
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CALL clearChecksum
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SETD.1 Whole
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CALL checksum
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CALL keepChecksum
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CALL clearChecksum
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CALL clearIndex
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smallLoop:
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SETD.0 SmallName
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SETD.1 Block
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SETD.2 Index
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LDA.2
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INCD.2
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LDB.2
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SWI osFileBlock
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BNQ smallDone
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CALL takeCount
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SETD.1 Block
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CALL checksum
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CALL stepIndex
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BRI smallLoop
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smallDone:
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SETD.0 BothText
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SWI osPrintString
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CALL printChecksum
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SETD.0 AgainstText
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SWI osPrintString
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CALL printKept
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SETD.0 NewLine
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SWI osPrintString
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CALL sameAsKept
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BNQ differ
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SETD.0 SameText
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SWI osPrintString
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BRI interleave
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differ:
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SETD.0 DifferText
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SWI osPrintString
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; ---- 4. Two files, alternately ----
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;
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; Block zero of the big file, then a block of the small one, then block zero of the big
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; file again. The two readings of the same block have to match. A system that remembered
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; the first file and did not notice the name had changed would hand back a block of the
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; wrong file in the middle, and then the right one again, so only the middle call would
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; be wrong - which is why this asks for the same block twice rather than once.
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interleave:
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CALL clearChecksum
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CALL readFirstBig
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CALL keepChecksum
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SETD.0 SmallName
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SETD.1 Block
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RSTA
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RSTB
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SWI osFileBlock
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CALL clearChecksum
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CALL readFirstBig
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CALL sameAsKept
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BNQ mixedUp
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SETD.0 InterleaveOk
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SWI osPrintString
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BRI moved
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mixedUp:
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SETD.0 InterleaveBad
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SWI osPrintString
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; ---- 5. A file that moves out from under the name ----
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;
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; The system has just been asked about the small file, so it is the one being remembered.
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; Renaming it has to throw that away: the blocks are still there and still hold the same
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; bytes, so a stale answer would work perfectly and be wrong.
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moved:
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SETD.0 SmallName
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SETD.1 OtherName
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SWI osFileRename
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BNQ noRename
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SETD.0 MovedText
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SWI osPrintString
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SETD.0 SmallName
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SWI osFileInfo
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CALL keepWhy
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SETD.0 OldNameText
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SWI osPrintString
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CALL printWhy
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SETD.0 NewNameText
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SWI osPrintString
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SETD.0 OtherName
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SWI osFileInfo
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CALL keepWhy
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CALL printWhy
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; ---- 6. The three ways of being told no ----
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missing:
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SETD.0 MissingName
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SWI osFileInfo
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CALL keepWhy
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SETD.0 MissingText
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SWI osPrintString
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CALL printWhy
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SETD.0 OtherName
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SETD.1 Block
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INIA 0xFF
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INIB 0xFF
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SWI osFileBlock
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CALL keepWhy
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SETD.0 PastText
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SWI osPrintString
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CALL printWhy
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SWI osExit
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noBig:
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CALL keepWhy
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SETD.0 NoBigText
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SWI osPrintString
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CALL printWhy
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SWI osExit
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noSmall:
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SETD.0 NoSmallText
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SWI osPrintString
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SWI osExit
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noRename:
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SETD.0 NoRenameText
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SWI osPrintString
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SWI osExit
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; ---- Routines ----
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; Block zero of the big file, into the running checksum.
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readFirstBig:
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SETD.0 BigName
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SETD.1 Block
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RSTA
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RSTB
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SWI osFileBlock
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BNQ readFirstDone
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CALL takeCount
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SETD.1 Block
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CALL checksum
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readFirstDone:
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RET
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; What the service just answered in DP3 becomes Left, which is what the checksum counts
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; down. Kept in memory rather than in a pointer because a CALL does not preserve one.
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takeCount:
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PSHD.3
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POPB
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POPA
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SETD.2 Left
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STA.2
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INCD.2
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STB.2
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RET
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; Adds the bytes at DP1 into the running checksum, as many of them as Left says.
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;
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; Two accumulators, each a byte wide, each throwing away what carries off the top. The
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; first is the sum of the bytes and the second is the sum of the first, so a byte that
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; arrives late counts for less than one that arrived early - which is what makes this
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; notice a block delivered out of turn.
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checksum:
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checksumLoop:
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LDA.1
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SETD.2 Fletch1
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LDB.2
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CCF
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ADD
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MVQA
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STA.2
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SETD.2 Fletch2
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LDB.2
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CCF
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ADD
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MVQA
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STA.2
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INCD.1
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; Left goes down by one, sixteen bits of it: a whole block is 256 bytes and a whole file
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; is more than one block, so a byte counter would not reach.
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SETD.2 Left
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INCD.2
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LDA.2
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BNA checksumLow
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DECD.2
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LDA.2
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DECA
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STA.2 ; Borrow out of the high byte.
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INCD.2
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INIA 0xFF
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STA.2
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BRI checksumTest
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checksumLow:
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DECA
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STA.2
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checksumTest:
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SETD.2 Left
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LDA.2
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INCD.2
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LDB.2
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OR ; Zero only when both halves are.
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BNQ checksumLoop
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RET
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clearChecksum:
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RSTA
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SETD.2 Fletch1
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STA.2
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SETD.2 Fletch2
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STA.2
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RET
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clearIndex:
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RSTA
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SETD.2 Index
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STA.2
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INCD.2
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STA.2
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RET
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stepIndex:
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SETD.2 Index
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INCD.2
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LDA.2
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INCA
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STA.2
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BNC stepIndexDone
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DECD.2
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LDA.2
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INCA
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STA.2
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stepIndexDone:
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RET
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; Puts the checksum aside so that a second one can be compared with it.
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keepChecksum:
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SETD.2 Fletch1
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LDA.2
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SETD.2 Kept1
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STA.2
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SETD.2 Fletch2
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LDA.2
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SETD.2 Kept2
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STA.2
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RET
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; Q is zero if the running checksum is the one that was put aside.
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sameAsKept:
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SETD.2 Fletch1
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LDA.2
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SETD.2 Kept1
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LDB.2
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XOR
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BNQ sameAsKeptDone
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SETD.2 Fletch2
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LDA.2
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SETD.2 Kept2
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LDB.2
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XOR
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sameAsKeptDone:
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RET
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printChecksum:
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SETD.2 Fletch1
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LDA.2
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SETD.2 Fletch2
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LDB.2
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SWI osPrintNumber
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RET
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printKept:
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SETD.2 Kept1
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LDA.2
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SETD.2 Kept2
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LDB.2
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SWI osPrintNumber
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RET
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; Why the last service said no. Q survives a CALL, which is the only reason this can be a
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; routine at all, but it does not survive the next SWI - so it is written down here and
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; printed later, with whatever has to happen in between happening in between.
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keepWhy:
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MVQA
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SETD.2 Why
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STA.2
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RET
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printWhy:
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RSTA
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SETD.2 Why
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LDB.2
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SWI osPrintNumber
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SETD.0 NewLine
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SWI osPrintString
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RET
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#Data
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#Base 0x2000
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BigName:
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"big.txt"
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SmallName:
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"small.txt"
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OtherName:
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"moved.txt"
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MissingName:
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"nothing.txt"
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BigIs:
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"big.txt is "
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BlocksText:
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" blocks
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"
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ReadText:
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"read "
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BlocksSumText:
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" blocks, checksum "
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StoppedText:
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", stopped with "
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BothText:
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"small.txt streamed is "
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AgainstText:
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", read whole is "
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SameText:
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"the same
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"
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DifferText:
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"DIFFERENT
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"
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InterleaveOk:
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"the same block twice with another file between: the same
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"
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InterleaveBad:
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"the same block twice with another file between: DIFFERENT
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"
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MovedText:
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"renamed small.txt
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"
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OldNameText:
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"the old name now answers "
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NewNameText:
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"the new name answers "
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MissingText:
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"a name that was never there answers "
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PastText:
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"a block past the end answers "
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NoBigText:
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"big.txt would not open, answer "
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NoSmallText:
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"small.txt would not read
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"
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NoRenameText:
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"it would not rename
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"
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NewLine:
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"
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"
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Index:
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0x00 0x00
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Left:
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0x00 0x00
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Fletch1:
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0x00
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Fletch2:
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0x00
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Kept1:
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0x00
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Kept2:
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0x00
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Why:
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0x00
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; One block, which is the whole point: the big file is four hundred times this.
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Block:
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#Reserve 0d256
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; And room for the small one all at once, so that the two ways of reading it can be
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; compared against each other.
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Whole:
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#Reserve 0d1024
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