The Z80 has 65536 I/O addresses, not 256. But just like its bigger
cousins, the x86 series from Intel, they are not directly accesible. If
I'm not mistaken you have to use the BC register to access ports above
0xff (of course you can also use it for the first 256 ports, but what use
would that be if there's a smaller and faster opcode for that?).
Thus, it would seem like this:
LD BC, 0x1234
LD A, 0x56
OUT [BC], A
Sorry, it's been some time since I did exensive Z80 programming so I may
have slipped with the example there, but I think the general idea should
be clear.
Greetz,
Martin 'Desert Mouse' Bethlehem.
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