On Sat, 5 Feb 2005 02:08:45 -0500, Daniel Phillips <[EMAIL PROTECTED]> wrote: > On Friday 04 February 2005 19:29, Lourens Veen wrote: > > On Friday 04 February 2005 23:21, Daniel Phillips wrote: > > > On Friday 04 February 2005 14:40, Timothy Miller wrote: > > > > You have a single chunk of logic that must iterate over pixel > > > > pairs. In order for it to know what to compute for one cycle, it > > > > must know the result of the previous cycle. If the computation > > > > takes more than one cycle, then it cannot iterate once per cycle. > > > > Latency becomes throughput. > > > > > > But why can't we solve this with an intermediate queue that holds > > > the same number of entries as stages in the iteration? > > > > Because the producer and the consumer are one and the same piece of > > logic > > I don't believe they are, and even if they were, you'd replicate it. > The problem Timothy is talking about (I think) has more to do with how > you factor the stages and move data between them.
No, Lourens is right. But you are correct in that some amount of replication can be done to compensate for that. > > > and it can only do one thing at a time (ie, read new input, > > process it, and put it into the queue, or read something from the > > queue, process it, and pass it on). > > Regards, > > Daniel > _______________________________________________ > Open-graphics mailing list > [email protected] > http://lists.duskglow.com/mailman/listinfo/open-graphics > List service provided by Duskglow Consulting, LLC (www.duskglow.com) > _______________________________________________ Open-graphics mailing list [email protected] http://lists.duskglow.com/mailman/listinfo/open-graphics List service provided by Duskglow Consulting, LLC (www.duskglow.com)
