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Flavioweb
Registered: Nov 2011
Posts: 463 |
Opcode Execution time != cpu cycles?
As writed at: http://www.atarimax.com/jindroush.atari.org/aopc.html#TIMES
"
Execution Times
Op code execution times are measured in machine cycles, one of which equals two clock cycles. Many instructions require one extra cycle for execution if a page boundary is crossed; ...
"
This mean that using a Cia timer setup to count system cycles and supposing to starting it just after an LDA #$00 instruction and reading it again (the timer value) after instruction execution, since LDA time is 2 machine cycles the timer value can be 4 (or better... to have a timer value of "0" we need to setup it for starting at 4)? |
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Flavioweb
Registered: Nov 2011
Posts: 463 |
Quote: Quoting Flavioweb
This mean that using a Cia timer setup to count system cycles and supposing to starting it just after an LDA #$00 instruction and reading it again (the timer value) after instruction execution, since LDA time is 2 machine cycles the timer value can be 4 (or better... to have a timer value of "0" we need to setup it for starting at 4)?
No, on LDA #$00 you always have 2 clock cycles and also the CIA will always count 2. Both, CPU and CIA use the same 1 MHz clock.
It should be noted that the CPU uses both edges (up and down) of the clock to do something, I think that's what they are trying to say.
Maybe that 1 "system cycle" are 2 edges, but 1 "clock cycle" is only 1 edge ?
So, if in a pal system with clock at 0,9875MHZ there are 987.500 "edges" per second, the cia can count "only" 493.750 "cycles" per second.
And the cpu to -execute- the LDA #$00 uses 2 -system- cycles that match at 4 "edges" or clock cycles...
correct? |
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chatGPZ
Registered: Dec 2001
Posts: 11386 |
Quote:Can we put that in the VIC article? :)
this stuff is explained right at the beginning of said article, *afaik* =)
other than that i recommend - especially for a beginner - to completely ignore the "machine cycle" vs "cpu cycle" thing. its the same in all practical situations anyway :) |
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Graham
Account closed
Registered: Dec 2002
Posts: 990 |
Quoting FlaviowebMaybe that 1 "system cycle" are 2 edges, but 1 "clock cycle" is only 1 edge ?
http://en.wikipedia.org/wiki/File:SquareWave.gif
Take a look at that graph: 0.0 to 1.0 is one cycle.
Quote:So, if in a pal system with clock at 0,9875MHZ there are 987.500 "edges" per second, the cia can count "only" 493.750 "cycles" per second.
No, there are 1 million "high" states per second, each of them followed by a "low" state. That's the definition of a "cycle": It repeats on and on.
Quote:And the cpu to -execute- the LDA #$00 uses 2 -system- cycles that match at 4 "edges" or clock cycles...
correct?
It's 2 clock cycles. The rest is pretty much irrelevant on the software side. What groepaz said: On 6502 machine cycles do not matter at all. You can completely ignore that something like that exists.
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Copyfault
Registered: Dec 2001
Posts: 478 |
Quote: Ok...
But CIA timer goes from 4 to 0 (3 to 0?) or from 2 to 0 (1 to 0?) during execution of LDA #$00 ?
Be careful with value '0' in timer regs. In theory, once a CIA timer is started it counts down from its inital value to zero. In reality the timer takes the init value again when reaching the 'zero state'. So it rather looks like this:
initial value e.g. = 4
at next clock cycle = 3
at next clock cycle =2
at next clock cycle = 1
at next clock cycle = 4
(in one shot mode, it stops here; otherwise it continues like this:)
at next clock cycle = 4
at next clock cycle =3
etc.
Confused me a lot when I had my first try with the CIA timers. |
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chatGPZ
Registered: Dec 2001
Posts: 11386 |
also that is completely irrelevant unless you are trying to use the timer value right before it wraps. i cant tell any practical situation where this is true right now to be honest :o) |
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JackAsser
Registered: Jun 2002
Posts: 2014 |
Quote: also that is completely irrelevant unless you are trying to use the timer value right before it wraps. i cant tell any practical situation where this is true right now to be honest :o)
When using a timer to stabilize a raster it can be confusing if you assume the value 0 exists... :) But yeah... not really relevant in most cases indeed. |
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Copyfault
Registered: Dec 2001
Posts: 478 |
Mainly I wanted to point this out because I think it helps a lot - at least it would've helped me back then ;)
In practice it could help to do a smart dejitter routine. This "zero state = init value"-thingie forced me to use the illegal asr-opcode ;) |
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chatGPZ
Registered: Dec 2001
Posts: 11386 |
or just start the timer one cycle later perhaps? =P |
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JackAsser
Registered: Jun 2002
Posts: 2014 |
Quote: or just start the timer one cycle later perhaps? =P
Totally off-topic of course, but since I formulated a problem and you posted a solution (the one we all use), it's not completely irrelevant knowing that the timer actually doesn't reach 0. :) |
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chatGPZ
Registered: Dec 2001
Posts: 11386 |
i have used timers for stable raster long before i knew about the wraparound bug though :) |
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