Hi,
You want to know the timing of an interrupt, is that correct?
(I have not read all the messages, so I might be wrong)
When an interrupt is received it takes, the 9S08 saves the CPU registers on the stack, except for reg H. That is it pushes A, X and CCR on the stack, 6 cycles.
It then pushes the PC on the stack, 2 * 2 cycles = 4 cycles.
It fetches the interrupt vector, that is 2 * 2 cycles or 4 cycles and modifies the PC.
So the interrupt takes 6 + 4 + 4 = 14 cycles to enter the interrupt routine.
To this must be added the time of the current instruction to be finished.
The slowest instruction is a version of DBNZ. It takes 8 bus clocks.
So the time for the interupt is somwere between 14 and 14 + 8 = 22 clock cycles.
Was that what you wanted to know?
I saw that you were discussing the SWI instruction.
Regards,
Ake