On Tue, 8 Feb 2005, Pete Turnbull wrote:
> Interesting -- this is an 11/23, yes? It presumably has 64KB of
> memory, or more. However, what a program can see is at most 56KB of
> addresses, and the remaining 8K is the I/O page. When you look at
> 165426 with ODT, you see the physical memory, but when the processor
> does it, any address above 157777 is mapped to the I/O page. To see
> what the processor would see while it's executing the bootstrap, you
> would need to look at 17765426. Unless it's one of the early KDF11-As,
> with only 18-bit ODT, in which case you should look at 765426 (17765426
> would actually give the same result, though, because of the way ODT
> works).
Ok, so I checked at 17765426 and this time I see what looks like code. In
fact, I dumped memory from 765400 through 765456:
_at_17765400/000761
765402/005721
765404/100406
765406/122737
765410/000240
765412/000000
765414/001002
765416/005007
765420/005741
765422/011102
765424/000000
765426/000167
765430/177360
765432/177404
765434/052700
765436/000010
765440/006300
765442/103376
765444/010061
765446/000006
765450/005000
765452/062700
765454/000005
765456/012761
The break point is at 765426, which means the actual break is at 765424,
and there's a 0 there. Aha! I think. Can someone less lazy (or who has
more time than I do now) disassemble this to see what it is? Is it RAM or
ROM? I can't seem to poke any words there.
Hmmmmmmmmm............
--
Sellam Ismail Vintage Computer Festival
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Received on Tue Feb 08 2005 - 23:23:08 GMT