XPost: comp.sys.sinclair   
   From: kr-lund@nogarbage.online.no   
      
   A936@hotmail.com wrote:   
   > Robert Newson wrote:   
   >   
   >>>identifying a specific device and the top 8-bits to select registers on   
   >>>the device.  The z80's instruction set also "prefers" 8-bit i/o ports.   
   >>   
   >>AFICR, the Z80 when using the C reg to define which port to use (on the   
   >>lower 8 address lines, as in IN A,(C) or OUT A,(C) ) also puts out the B reg   
   >>on the top 8 address lines - the zx81 uses this to scan the keyboard with   
   >>the B reg activating the "current row".   
   >>   
   >>When the immediate mode version is used (as in IN A,80 or OUT A,80) I'm not   
   >>sure, but somewhere lurking round my memory suggests, that the current   
   >>contents of the A reg is used for the top 8 bits - can anyone confirm/deny   
   this?   
   >   
   >   
   > Yes the A register is placed in the top 8 bits.  This is compatible   
   > with   
   > the 8080 and is the reason I said the z80 "prefers" 8-bit i/o ports.   
   >   
   > The other reason is that the block i/o instructions use the B register   
   > as a counter and also place the count in the top 8 bits for each i/o   
   > step.   
   > This may or may not be useful for 16-bit i/o depending on whether   
   > this is appropriate for the device in question and on whether the   
   > designer had the wherewithall to make use of this feature.   
   >   
   >   
   >>None of my (rather scant) Z80 programming books describe the ports as   
   >>"16-bit" addressed, only 8-bit with the port number either as immediate data   
   >>or in the C reg.   
   >   
   >   
   > Despite many references on the internet that this is undocumented,   
   > it is not.  It is very much documented in every data book I've ever   
   > seen   
   > including the earliest one I have -- a 1979 data book from Mostek.   
   > The confusion, I assume, comes from early books (ie not data books)   
   > written by authors familiar with the 8080 who failed to do their   
   > homework   
   > and therefore didn't realize that the i/o space was 16-bit on the z80   
   > :-)   
   >   
      
   Z80   
      
   Note that 16bit I/O is not very useful on the TRS-80 I/III/4 machines   
   and clones as only the lower 8 bit are decoded. But one could use it for   
   special self-made hardware on those 8bit I/O addresses that were free.   
      
   I think this is not mentioned yet. The actual cycle that Z80's I/O   
   creates is slower than that for memory access. This makes it easier to   
   interface.   
      
   --   
   Knut   
   (delete 'nogarbage.' for email)   
      
   --- SoupGate-Win32 v1.05   
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