The Z8000 included the Zilog (later Exxon Office Systems) Zeus office automation Unix boxes... They were SystemIII and used by the US Government's Internal Revenue Service through 1993 or so. They later adopted Pyramid multiprocessor RISC boxes. Bill --- Bill Gates is a Persian cat and a monocle away from being a villain in a James Bond movie -- Dennis Miller bpechter@shell.monmouth.com|pechter@pechter.dyndns.org

One Circuit Cellar project was a co-processor card for the IBM PC containing a Z8000. I toyed around a while with building an S100 bus processor card for Z8000 (long after the both the chip and systems using the bus were no longer in general production.) Never came to anything. I still have to processor technical manual on my shelf. It's an interesting chip, but was a bit slower than it needed to be. Eric

It was thus said that the Great "Fred Cisin (XenoSoft)" once stated: It doesn't have to be at 0, but it does have to be the last executable instruction other than NOP (which is $00). Yes. 8-) -spc (It only took me two hours because I've never programed the 8080/Z80 before ... )

Iggy Drougge wrote: Olivetti made one IIRC it was the M21 - still looking for one here in France..... -- HBP

When I think of MOS, I think of a multi-user MS-DOS clone. We had a server running it on a small (less than a dozen systems) network in the late 80's. I no longer remember the protocal or NIC type used but we had a bunch of Z-248's connected to this machine, which I believe was a '386, using coax cable. It replaced a single CP/M system for keeping our troublecall database. Jeff -- Home of the TRS-80 Model 2000 FAQ File http://www.cchaven.com http://www.geocities.com/siliconvalley/lakes/6757

Jim Battle wrote: <snipped solution> No, RST 0 will push the return address. There are quite a few solutions to this. Here are some given in the June 1979 issue of PCW: For the purists who think that the code should start at 0: 0000 21 06 00 LXI H,0006 0003 54 MOV D,H 0004 5C MOV E,H 0005 F9 SPHL 0006 D5 PUSH D 0007 E9 PCHL The first instruction sets the HL registers to 0006 (00 in H, 06 in L) and the next two set the D and E registers to zero using the data (00) in H. The stack pointer SP is then loaded with 0006 from HL. 'Pushing' the DE registers in the next instruction writes zeros into locations 0004 and 0005 and the stack pointer is reduced by two to 0004. The next instruction, PCHL, means 'jump to the address stored in HL' which, since HL contains 0006 means that the execution continues with the PUSH D instruction. Successive Pushes write pairs of zero bytes _down_ through memory, past location 0000 to FFFF and on until, after 32767 pushes, the only two bytes in memory which are not zero are the PUSH D and PCHL instructions themselves. The 32768th PUSH overwrites these last two and the whole of memory is set to zero. The most popular of the shortest possible solutions was this: FFFA 31 FC FF LXI SP,FFFC FFFD CD FD FF CALL FFFD The first instruction sets the stack pointer to FFFC, the byte before the next instruction. The CALL instruction which follows calls itself at location FFFD pushing the return address, which happens to be 0000, onto the stack. With successive CALLs the stack pointer works downwards through memory writing pairs of zero bytes. After the 32767th CALL only two bytes, FFFD and FFFE, remain unaffected so that the last CALL, which overwrites these last two bytes, jumps to location 00FD.

Iggy Drougge wrote: No, they're straight from the magazine, from a time when I imagine many hobbyists would be interested. Now I think assembler only interests those of us who still have to count clock cycles for a living! Regards, Paul