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. > 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)
