Re: [hatari-devel] DSP bug: need more explanations

 - r0 is, in fact, used: r0 = 0 1. iteration r0 = 0 old_r0 +1 => r1 r1 = 1 2. iteration r0 = 1 old_r0 + 1 => r1 r1 = 1 3. iteration r0 = 2 old_r0 + 1 => r1 r1 = 2 ... something like that? Yes, but for the third iteration, if there was pipeline effect, I think that at the end of the movem, R0 = 1 and not 2 (as R1 in the second iteration = 1 and writes its value to p:\$118) > - you're using 2 program words, i.e. it could somehow (don't ask me how) slow down the whole process and r0 gets the right value. what if you replace movem with, say, move r2,r0? Yes, great idea, I test immediatly (I'll use X1 to be sure there's no pipeline effect here ;) Hard problem for now ;) Regards Laurent   Le 04/01/2015 23:14, Miro Kropáček a écrit : Ah so, now I get it. On real hardware, I get r1 = 2 r2 = 1 r3 = 1 Yeah, this makes sense, as discussed before. On real hardware, I get : a = \$20 [...]  From my point of view : After the first loop, R0 = 0 and p:\$118 = 1 At the beginning of the 2nd loop : r0 goes from 0 to 1 (the value in p:\$118) , but may suffers the pipeline effect Why don't you initialize r0 to some non-zero value, like \$0100 for example?   So, the next lua (r0)+,r1 should take R0 = 0 and R1 = 1 The final result with a pipeline effect should be 1, not 20 Which means that the   lua (r0)+,r1  doesn't suffer the pipeline effect in this case.  Three things I could think of: - you're using 2 program words, i.e. it could somehow (don't ask me how) slow down the whole process and r0 gets the right value. what if you replace movem with, say, move r2,r0? - you're tinkering with 'do' loop -- there are some restrictions for end of the loop situations, perhaps worth checking, too - r0 is, in fact, used: r0 = 0 1. iteration r0 = 0 old_r0 +1 => r1 r1 = 1 2. iteration r0 = 1 old_r0 + 1 => r1 r1 = 1 3. iteration r0 = 2 old_r0 + 1 => r1 r1 = 2 ... something like that? Just from the top of my head, I can be terribly wrong, as usual ;) --  MiKRO / Mystic Bytes http://mikro.atari.org

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