AnachronautandClaude Opus 5
00d896e3e7
Break left DP1 and DP2 alone, so what a stop printed for them was whatever the shell happened to have left there - which is a CosmOS address, which moves whenever CosmOS is touched at all. The recorded output had to be taken again four times in one day's work, every time for a value that is not this program's and that nothing should ever depend on. It sets all three of the pointers it owns now, rotated between the two stops so that every one of them visibly changes, the way A and B already did. DP3 stays as the system left it: it is where the program was entered, which is the one thing worth seeing here that this program did not choose, and it is steady because it is this program's own base. A demonstration of what the registers were should show registers somebody chose. Then every line of the record is being asserted rather than merely observed, and a reader can tell which is which. Checked both ways: sixty four bytes added to CosmOS's data no longer moves it, and reading the frame at the wrong offset still fails it. cosmosRun and cosmosMonitor move because Break is sixteen bytes longer and both of them list the disk it sits on. Co-Authored-By: Claude Opus 5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_01E2JrLzFvuFX9fgi1LDRjrW
84 lines
2.7 KiB
NASM
84 lines
2.7 KiB
NASM
; Stopping a program to look at it.
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;
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; SWI osBreak shows every register as this program had them, waits for a key, and carries
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; on. It is two bytes and it fires every time it is reached.
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;
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; WHY IT IS AN INSTRUCTION RATHER THAN SOMETHING SET FROM OUTSIDE. A breakpoint that was
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; poked into a running program would have to overwrite an instruction, and then putting that
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; instruction back in order to continue is the same act as disarming the breakpoint. Firing
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; a second time would mean stepping over the restored instruction and putting the breakpoint
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; back behind it, and this machine cannot step one instruction. Nothing is overwritten here,
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; so there is nothing to restore and nothing to re-arm.
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;
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; The price is that it is part of the program. A build with breakpoints in it has different
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; addresses from a build without, which is the same bargain every machine makes that has a
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; break instruction.
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;
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; Correct output is two stops. A and B differ between them, the addresses differ, and the
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; Stack Pointer differs too, because the second one is inside a subroutine and a call has
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; put ten bytes down by then.
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;
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; EVERY POINTER IS SET BEFORE EACH STOP, and the three this program owns are rotated
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; between the two so that all of them visibly change.
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;
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; This program used to leave DP1 and DP2 alone, and what a stop then showed for them was
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; whatever the shell happened to have left there. That is a real thing about the machine -
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; a program is handed the pointers as it finds them - but it is not this program's to
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; demonstrate, and it made what this program prints depend on where CosmOS's code happens
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; to sit. The recorded output had to be taken again four times in one day's work, every
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; time for a value nothing should ever depend on.
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;
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; A demonstration of what the registers were should show registers somebody chose. Then
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; every line of it is being asserted rather than merely observed.
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;
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; DP3 is deliberately still the system's. It is where the program was entered, which is the
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; one thing here worth seeing that this program did not choose, and it is steady because it
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; is this program's own base.
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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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SETD.0 Banner
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SWI osPrintString
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INIA 0d17
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INIB 0d34
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SETD.0 Marker
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SETD.1 Banner
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SETD.2 DoneText
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SWI osBreak
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; The second stop is inside a subroutine, so that the Stack Pointer is visibly not where
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; it was: a call puts ten bytes down before this one gets there.
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CALL deeper
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SETD.0 DoneText
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SWI osPrintString
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SWI osExit
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deeper:
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INIA 0d68
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INIB 0d85
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SETD.0 Banner
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SETD.1 DoneText
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SETD.2 Marker
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SWI osBreak
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RET
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#Data
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#Base 0x2000
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Banner:
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"two stops, and what the registers were at each
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"
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Marker:
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"marker"
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DoneText:
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"carried on to the end
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"
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