Edwin Blink wrote:
> From: Geoff Winkless <[EMAIL PROTECTED]>
>
>> Without wishing to appear stupid (yeah, I know, too late, haha), I'm
>> still not clear why that's a memory access.
>
> It's not. The delay is added there for convenience. But in fact it
> will delay the next
> opcode fetch (unless it is in rom but read below).
>
>> Shouldn't that round up to 4,4,8,8,5 if in ROM with display
>> contention
>> and 4,4,4,8,5 with RAM contention?
>
> The delay shifts to the next (HL) cycle on next LDIR.
Whuh?
> the RAM can be accessed once every 8T during display(PAPER for Ales)
> contention and once every 4Ts during RAM contention
But there will be no RAM access for 8 t-states after the ,5 - the 4,4 will
take 8 t-states, so the RAM will be available again.
If what you're saying is that actually everything is broken down into chunks
of 8-t-states _regardless_ of whether there's a memory access, then that's a
bit more understandable...
_However_, what that actually means is:
If instructions in []s are ROM-based (or internal) and in {}s are RAM-based
(therefore must be 8-state aligned):
[4],[4] (one block) -- 8 cycles
{3} (one block) -- 8 cycles
{5} (one block) -- 8 cycles
[5] + [3 of 4] (one block) -- 8 cycles
[remaining 1 of 4] + [4] + {3} (one block) -- 8 cycles
{5} (one block) -- 8 cycles
[5] + [3 of 4] (one block) -- 8 cycles
[remaining 1 of 4] + [4] + {3} (one block) -- 8 cycles
This suggests that after the initial round, it will in fact only take 24
cycles per byte.
Actually, it's more like this:
[4],[4] (one block) -- 8 cycles
{3} (one block) -- 8 cycles
{5} + 3 of [5] (one block) -- 8 cycles
[2] of [5] + [4] + [2] of [4] (one block) -- 8 cycles
remaining [2] of [4] + {3} (one block) -- 8 cycles
but that's effectively the same.
If you go back to "each display-contended access takes exactly 8 cycles"
then you have
[4],[4]
{3} -- 8 cycles
{5} -- 8 cycles
[5]
or 29 cycles.
So which is it? Is it RAM availability is in 8-cycle blocks or each RAM
access takes exactly 8 cycles?
Or is it worse than that?
Please understand that I'm not trying to be awkward; I just don't comprehend
the logic.
Cheers!
Geoff
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