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Hypnosis
Registered: Mar 2015
Posts: 36 |
Cycles to get into IRQ routine
How many cycles does it take from an IRQ trigger to the start of the IRQ routine, assuming the CPU finished the instruction when it triggers? It should be around 6-9 cycles but I'm not sure how many. |
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lft
Registered: Jul 2007
Posts: 369 |
Just to establish some common terminology:
Latency: The number of cycles from an interrupt until the first instruction in the interrupt handler starts to execute.
Jitter: The random variation in latency, i.e. the difference between minimum latency and maximum latency.
If minimum latency is equal to maximum latency, the amount of jitter is zero. Thus, if we observe e.g. four different timer values, there is a jitter of three cycles.
Returning to the topic, here's a recapitulation:
The 6502 normally checks for interrupts at the end of the second-to-last cycle of an instruction. The minimum latency is therefore 2 + 7 cycles: two for the two last cycles of the current instruction, and seven to handle the interrupt (push on stack etc.)
If we get the interrupt while executing a stream of nops, as in the double-irq method, the maximum latency is 1 + 2 + 7 cycles: one for the last cycle of the first nop (we just missed the check!), and then as above.
The longest legal instruction is seven cycles (e.g. asl abs,x). If an interrupt happens during the final cycle of the previous instruction, the latency is therefore 1 + 7 + 7 cycles.
So far, we have a jitter of (1 + 7 + 7) - (2 + 7) = 6 cycles.
Now we add the special case discussed in this thread, where a non-page-crossing branch "sticks" to the subsequent instruction (the branch target). Maximum latency is now 3 + 7 + 7, and the jitter becomes 8 cycles.
With illegal opcodes, the longest instruction is 8 cycles. If we also consider the delayed IRQ on rasterline 0 as latency, the total jitter becomes (1 + 3 + 8 + 7) - (2 + 7) = 10 cycles.
That is, we can expect 11 different timer values.
Did I get that right? |
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Flavioweb
Registered: Nov 2011
Posts: 463 |
Lft: yep, you're right. |
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lft
Registered: Jul 2007
Posts: 369 |
No, hang on. For branch instructions, the interrupt flag is actually checked at the end of the first cycle (but not at the end of the second cycle). So for the worst case, we're only stuck with the last two cycles of the branch. (1 + 2 + 8 + 7) - (2 + 7) = 9 cycles jitter, 10 different timer values. |
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Flavioweb
Registered: Nov 2011
Posts: 463 |
Yes.
If second cycle of a taken branch opcode, is executed on cycle zero of a rasterline (first cycle of branch op is on last cycle of previous line), then cycles added to wait for current opcode to finish are 2 from taken branch, plus next opcode cycles.
So, if we have branch opcode + 7 cycles opcode, cpu wait for 2 cycles from branch + 7 cycles from next opcode.
If we are on raster line 0, and first cycle of branch is on first cycle of the line, irq is triggered on cycle 1 so our first irq opcode is far 1 + 2 + 7 + 7 = 17 cycles from first cycle of the line. |
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Copyfault
Registered: Dec 2001
Posts: 478 |
Quoting lftJust to establish some common terminology:
Latency: The number of cycles from an interrupt until the first instruction in the interrupt handler starts to execute.
Jitter: The random variation in latency, i.e. the difference between minimum latency and maximum latency.
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Thanks for giving a proper definition of jitter - I'd even suggest to say jitter value as jitter refers more to the phenomenon itself ;).
To be honest I have been using the term jitter (value) in two different ways: sometimes for the delay until the irq handler starts, which makes sense only for a fixed situation, on other occasions for the difference between the maximal and the minimal occuring latency, which makes sense only when looking at a complete set of situations (e.g. all possible situations).
To distinguish between them I also used current jitter value and maximal jitter value resp., but your definitions suit our needs way better.
Back to the actual jitter values: my tests give
Set of situations jitter value/differnt timer values
-------------------------------------- ----------------------------------
no illegals, raster irq outside line=0 7/8
illegals, raster irq outside line=0 8/9
no illegals, raster irq on line=0 8/9
illegals, raster irq on line=0 9/10
Missing in the list above is the Set of "NMI interrupts IRQ"-situations. Did anyone already have a look at these cases? |
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Hypnosis
Registered: Mar 2015
Posts: 36 |
I revisited this topic since I seemed to miscalculate when I was implementing a timer interrupt. I single stepped in Vice to see where the cycles ended up. I have a stable raster interrupt firing a timer A interrupt while executing nop instructions. I use timer B to measure cycles.
What happens in Vice is that
1. timer A wraps around
2. one nop is executed
3. another nop is executed
4. next instruction is now the nmi interrupt's first instruction
What's confusing is that there are 9 cycles between step 3 and 4. If so, why is the first nop executed? |
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lft
Registered: Jul 2007
Posts: 369 |
Possibly there is a delay before the CIA generates the interrupt. I believe the old and new CIA versions differ in the details here. |
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Flavioweb
Registered: Nov 2011
Posts: 463 |
Seven cycles are used to "setup things" and then go to irq first opcode.
Your code seems that trigger the irq on the second cycle of first nop, so the next opcode will be executed (next nop in this case).
So: irq triggered on second nop cycle, plus 2 cycles of next nop, plus seven cycles of irq "setup code": the first cycle of your irq code is executed 10 cycles after irq is triggered...
Edit: the irq is triggered on second nop cycle, probably because you are using an old cia, where irq is triggered 1 cycle "after". Maybe if you use a new cia, only first nop will be executed... |
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Hypnosis
Registered: Mar 2015
Posts: 36 |
Thanks for the input. I'll try the other CIA. This could be it. |
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Hypnosis
Registered: Mar 2015
Posts: 36 |
Yep. I get the expected behavior using the new CIA. |
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