Z-80 code question about a loop that depends on the contents of the refresh register
ard.p850ug1 at gmail.com
Wed Dec 14 12:36:16 CST 2016
>> I can't find any information on what the MSB is set to when the
>> accumulator is loaded with R, and what the Sign flag is set to. The
>> datasheet says the Sign and Zero flags are changed by the instruction.
>> If either of these flags are set, the routine enters an infinite loop.
> Tony Duell pointed that out too. I suspect that this code works for
> machines that existed at the time, with 16K DRAMs, and might or might
> not have worked on later machines. A quick scan of InfoWorld links
The type of RAM shouldn't matter (provided it works properly...)
The R register is an 8 bit register. It can be loaded from the accumulator
and read back into the accumulaotr (the LD A,R instruction you see here).
The bottom 7 bits of R are incremented (with wraparound from 111 1111 to
000 0000) after every instruction. The top bit is left alone.
At certain times, the RFSH/ signal is asserted by the Z80, At this point
all 8 bits of the R register appear on the bottom 8 bit of the address bus.
(As an aside, the I register, the interrupt vector register, appears on the
top half of the address bus).
Ths incrementing 7 bit address is useful for refreshing those DRAMs that
need 128 refresh cycles (a so-called '7 bit refresh'). There is no reason you
have to use it for that, even if you have DRAM (you could make your own RAM
controller, with its own refresh generator, in fact I would guess a lot of S100
RAM boards did, so they could be used with CPUs other than the Z80).
The behaviour of the R register is, of course, unchanged whether you use
it for RAM refresh or not.
It is my guess (without looking at the datasheets) that as the LD A,R
instruction has been said to affect the sign and zero flags that these
behave in the obvious way. Z is set if the value loaded is zero. The
sign bit is set if the high bit of R is set.
If that guess is right then there is a problem. If the high bit of R is set
then the routine as given will go into an infinite loop. Remember the automatic
incrementing of R is only on the bottom 7 bits.
What I need to do is find exactly how the flags are set by the
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