Jim Brain wrote: > Unfortunately, it does not work that way. I understood that the $01/$1 > registers are present in every bank, not just bank $0. Perhaps that is a > wrong assumption. As such, I also decode them to each bank. (I think it > has to be that way, for how else would you move from execution in bank 1 > to bank 0 if the registers only appear in bank 0.) So, a write to any > bank: $0/$1 will save new values in the banks (though they will not be > used until the developer moves back into emulation mode). Yes, the registers need to present in every bank. Regards, Michau.Received on 2020-05-29 21:33:48
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