Re: Native /// bootstrapping?

[dempson@actrix.gen.nz (David Empson)]

schmidtd <schmidtd@my-deja.com> wrote:

> On Feb 23, 5:57 pm, lyricalnan...@dosius.ath.cx (Steve Nickolas)
> wrote:
> > I'm trying to figure out how the SOS bootloader loads sos.kernel, and I'm
> > not really getting it (although apparently the bootloader flips the system
> > up to the topmost memory bank).
> 
> Jeppson has a pretty good discussion of the boot sequence:
> "Boot Sequence: On power-up, or after control-reset, the boot process
> begins in ROM#1 (ROM#2 doesn't yet exist). Low-level diagnostics are
> performed. Then block 0 is read from the disk in the built-in drive.
> This is the SOS boot code and is present on every disk that has been
> formatted by the System Utilities program. It must be present for a
> successful boot. It consists of one block of "absolute" code and is
> loaded into the computer at $A000, where it begins to run.
> 
> The boot code begins by locating and switching in the highest bank of
> RAM. Then it goes back to the disk and loads in five more blocks
> (blocks 1..5). These are placed in $A200..ABFF. Block 1 currently
> contains all zeros; blocks 2..5 are the disk directory. The boot code
> then scans the directory and locates SOS.Kernel, which it loads into
> memory at $1E00..73FF."
> http://apple3.org/Documents/Magazines/AppleIIIBits.html

Drive-by curiosity makes me ask...

How does that tally with the original layout of ProDOS 5.25" boot disks,
which have the ProDOS boot loader in block 0 and the SOS boot loader in
block 1?

Is there some trick the Apple /// is doing to recognise valid code in
block 0 and if it can't find it, trying block 1 instead, or is part of
the ProDOS boot block actually a SOS stub loader (at some offset into
the block) which fetches and executes block 1?

The ProDOS boot block starts with an $01 byte, which is not a valid 6502
instruction.

-- 
David Empson
dempson@actrix.gen.nz