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Date: Sun, 17 Apr 2011 02:05:47 -0700 From: Stephane Eranian <eranian@...gle.com> To: Robert Richter <robert.richter@....com> Cc: Peter Zijlstra <peterz@...radead.org>, Ingo Molnar <mingo@...e.hu>, LKML <linux-kernel@...r.kernel.org> Subject: Re: [PATCH 4/4] perf, x86: Fix event scheduler to solve complex scheduling problems On Sun, Apr 17, 2011 at 1:44 AM, Robert Richter <robert.richter@....com> wrote: > On 16.04.11 11:52:54, Stephane Eranian wrote: >> On Fri, Apr 15, 2011 at 5:27 PM, Robert Richter <robert.richter@....com> wrote: > >> > event code counter failure possible solution >> > >> > 1) 0x043 PMC[2:0] 0 1 >> > 0x02E PMC[3,0] 3 0 >> > 0x003 PMC3 FAIL 3 >> > >> I am not sure I understand this failure case. If I recall >> the scheduler looks at the weight of each event first: >> >> weight >> 1) 0x043 PMC[2:0] 3 >> 0x02E PMC[3,0] 2 >> 0x003 PMC3 1 >> >> Then, it schedules in increasing weight order. So it will >> schedule weight 1, 2, 3. For weight 1, it will find counter3, >> for weight 2, it will take counter0, for weight 3, it will >> take counter 1 given 0 is already used. >> >> Or am I reading your example the wrong way? > > No, I have to admit this one was taken out of my mind and I picked > a wrong example for the special problem I was thinking about. The > above works as you described because of the constraint's weights. > > I don't have an example with existing constraints, but consider the > following (theoretical) one: > > counter: 3210 Failure: Solution: > > e1 xx xo ox > e2 xx xo ox > e3 x x o x o x > e4 x x x F x o > > The special with the above is that two events (e1 and e2) must be > rescheduled to schedule e4. This means that swapping counters of only > one already scheduled event is not sufficient. A counter of a third > event must be freed, this counter is then taken by the second event. Ok, I understand this one better now. But you admit yourself, you made it up. Is it even possible with Fam15h? The thing is, I think, this is not a catastrophic problem. What will happen is that e4 cannot be scheduled with e1, e2, e3. In case the events are not in the same group, e4 will eventually get a chance to be scheduled due to round-robin on the event list. In case, the events are all in one group, then you won't be able to add e4 to the group because of group scheduling sanity check on creation. So either you have to shuffle your event groups or you will incur multiplexing. It's not like e4 will be created but it will nerver be able to count anything. >> The fact that counter have overlapping constraints >> should not matter. In fact this is what happens with >> counters without constraints. > > An event set containing constraints with following conditions is > problematic: > > (c1->weight <= c2->weight && (c1->cmask & c2->cmask != c1->cmask)) > > Basically this means both constraints do not overlap completly. You > can't select then the correct counter without knowing the events with > higher weights that will be scheduled later. > > -Robert > > -- > Advanced Micro Devices, Inc. > Operating System Research Center > > -- To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@...r.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
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