Hello Ruud, * On Thu, Oct 26, 2017 at 06:24:14PM +0200 Ruud@Baltissen.org wrote: > The adder adds one to the original address. Why? When the 8088 > addresses the the 0Fxxxxh area, say bank 15, it addresses the ROM. > But when addressing 0Fxxxxh, BP0..3 are all zero. But what happens > if the 8088 addresses the 0Exxxxh area, BP0..3 are all one. Or in > other words: bank 15! I do not know that card nor that Computer, but I would believe that the reason might be simple: The Z80 card on the C64 does the same. That's because in $0000-$00FF, the 6502 has its zero page. In $0100-$01FF, it has its (hard-coded) stack. For the 8080/Z80, there is the interrupt table at this location. For the 8088, it is the same, although the interrupt table is differently organized w.r.t. the 8080/Z80. The interrupt table would conflict with the ZP and the stack. By adding 1, you make sure that this conflict does not occur anymore. Regards, Spiro. -- Spiro R. Trikaliotis http://www.trikaliotis.net/ Message was sent through the cbm-hackers mailing listReceived on 2017-10-26 18:01:12
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