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Atari coding BBS
| Re: DSP extern memory extra cycles |
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Posted by: Sallafranque
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Jan,06.2012-01:32
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Hi,
Thanks a lot for your 2 answears.
This confirms what I understood about this external memory access.
I did some coding about cycles in the DSP code, but I missed this special point (multi extra access to external memory cost a 1 cycles waitstate after the first access).
It'll be quite easy to add this to Hatari now, I'll do it soon.
> And don't forget, some addressing modes ((r0)-,IIRC, for example) add additional cycle!
I've already implemented the special EA extra cycles into the dsp code of hatari.
I still have 3 little questions on this subject, just to be sure everything is understood :
1) In the table A-14 page A-232 of the DSP56001UM-A3 pdf, I can read the following accesses : L: XY: declined for
INT, INT
INT, EXT
EXT, INT
EXT, EXT
Is it only a typo in the doc (I thought that a L:access could only concern INT,INT or EXT,EXT access, as the same address is used for both X: and Y: access).
I'm OK for X or Y (upper address can be INT and lower address can be EXT for example)
Am I wrong ?
2) let's have a look at the move reg1,X:adr1 reg2,Y:adr2 instruction.
Let's consider adr1 and adr2 are exernal memory access.
According to the Table A-7, a X:Y: access is ea+axy cycles.
According to the Table A-14, axy is 2 + wx + wy cycles.
Our memory is 0 waitstate, but because of the bus, writing to X and then Y will cost 1 Waitstate.
So this instruction will execute in move cycles + ea cycles + 2 + 1 ?
3) Let's have a look at JMP (Table A-11)
According to the doc, "All one-word jump instructions execute TWO program memory fetches to refill the pipe-line, which is represented by the ‘‘+(2 ∗ ap)’’ term."
A JMP is 4 + JX cycles
Jx is ea + (2*AP) cycles
AP = 0 if internal memory or wp if external memory (0 waitstate for us)
Let's say our JMP istruction is located in P internal memory at address $50
and R2 is in internal memory too.
$50 JMP R2
The total cycles is 4
Now, let's say the JMP is in P external memory and R2 is also in external memory.
The total cycles should be :
4 (jmp) + 0 (ea) + 2*0 (ap) ?
or
4 (jmp) + 0 (ea) + 2*0 (ap) + 2 (3 external access to memory - 1) ?
Thanks for your patience,
Regards
Laurent
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