On Mon, 2007-11-06 at 13:58 +0200, Patrick McHardy wrote: > Thats not true. Assume PSL has lots of packets, PSH is empty. We > fill the PHL queue until their is no room left, so the driver > has to stop the queue.
Sure. Packets stashed on the any DMA ring are considered "gone to the wire". That is a very valid assumption to make. > Now some PSH packets arrive, but the queue > is stopped, no packets will be sent. > Now, you can argue that as > soon as the first PHL packet is sent there is room for more and > the queue will be activated again and we'll take PSH packets, _exactly_ ;-> > so it doesn't matter because we can't send two packets at once > anyway. Fine. i can see your thought process building - You are actually following what i am saying;-> > Take three HW queues, prio 0-2. The prio 2 queue > is entirely full, prio 1 has some packets queued and prio 0 is > empty. Now, because prio 2 is completely full, the driver has to > stop the queue. Before it can start it again it has to send all > prio 1 packets and then at least one packet of prio 2. Until > this happens, no packets can be queued to prio 0. The assumption is packets gone to the DMA are gone to the wire, thats it. If you have a strict prio scheduler, contention from the stack is only valid if they both arrive at the same time. If that happens then (assuming 0 is more important than 1 which is more important than 2) then 0 always wins over 1 which wins over 2. Same thing if you queue into hardware and the priorization is the same. cheers, jamal - To unsubscribe from this list: send the line "unsubscribe netdev" in the body of a message to [EMAIL PROTECTED] More majordomo info at http://vger.kernel.org/majordomo-info.html
