On 13 Aug, Don Lewis wrote: > I managed to figure out a bit more of what is going on. > > The following code in sched_choose() pulls the selected kse off the run > queue. > > ke = kseq_choose(kseq); > if (ke) { > #ifdef SMP > if (ke->ke_ksegrp->kg_pri_class == PRI_IDLE) > if (kseq_idled(kseq) == 0) > goto restart; > #endif > kseq_runq_rem(kseq, ke); > ke->ke_state = KES_THREAD; > > if (ke->ke_ksegrp->kg_pri_class == PRI_TIMESHARE) { > CTR4(KTR_ULE, "Run kse %p from %p (slice: %d, pri: %d)", > ke, ke->ke_runq, ke->ke_slice, > ke->ke_thread->td_priority); > } > return (ke); > } > > > At some later time, setrunqueue() gets called for this thread, which > calls sched_add(), which calls sched_add_internal(), which executes the > following code frament. > > class = PRI_BASE(kg->kg_pri_class); > switch (class) { > case PRI_ITHD: > case PRI_REALTIME: > ke->ke_runq = kseq->ksq_curr; > ke->ke_slice = SCHED_SLICE_MAX; > ke->ke_cpu = PCPU_GET(cpuid); > break; > case PRI_TIMESHARE: > if (SCHED_CURR(kg, ke)) > ke->ke_runq = kseq->ksq_curr; > else > ke->ke_runq = kseq->ksq_next; > break; > [snip] > kseq_runq_add(kseq, ke); > kseq_load_add(kseq, ke); > > > Because the thread is a CPU hog, it gets put on the next run queue, even > though it hasn't exhausted its current slice, which means that it has to > wait for all the other CPU hogs to get a turn at the CPU before it can > execute again. > > I don't know how to fix this problem. I think the desired behaviour > would be for the kse to be restored to its previous location on the run > queue. The best idea that I can come up with is for sched_add(), sched_add_internal(), kseq_runq_add(), and runq_add() to grow another parameter that would tell them whether to prepend to the beginning of the run queue or append to the end. If setrunqueue() detects that TD_IS_RUNNING(td) is true, it would pass the flag to sched_add() that would cause the thread to be added to the beginning of the queue. I don't know if this is appropriate in the PRI_ITHD and PRI_REALTIME cases, or if we want to continue to to round-robin. Comments?Received on Sat Aug 14 2004 - 03:23:34 UTC
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