Messages in this thread | | | Subject | Re: [PATCH] perf: Add a new sort order: SORT_INCLUSIVE (v4) | From | Peter Zijlstra <> | Date | Tue, 27 Mar 2012 21:38:18 +0200 |
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On Tue, 2012-03-27 at 11:09 -0700, Arun Sharma wrote: > On 3/24/12 7:14 PM, Frederic Weisbecker wrote: > > >> The other problem in branch stacks/LBR is that they're > >> sampled branches. Just because I got a sample with: > >> > >> a -> b > >> b -> c > >> > >> doesn't necessarily mean that the callchain was a -> b -> c. > > > > Not sure what you mean. If you have a -> b, b -> c in single > > LBR sample it means you got a -> b -> c. > > > > I was going by Stephane's commit message here: > > http://article.gmane.org/gmane.linux.kernel/1236999 > > > Statistical sampling of taken branch should not be confused > > for branch tracing. Not all branches are necessarily captured > > Stephane, could you please explain if the 16 filtered branches in LBR > are guaranteed to be from a given callchain to the leaf function? My > understanding is that it's not. > > callchain1: a -> b -> d -> e (sample a->b) > callchain2: a -> c -> b -> f (sample b->f) > > on PMU interrupt can we end up with: > > b -> f <- top of stack > a -> b > ... > > even though a -> b -> f can never happen in the actual program flow?
Right, so the LBR is a queue not a stack. A program like:
foo() { bar1(); bar2(); }
will, using the lbr, look like: foo->bar1->bar2 (if you filter returns), or foo->bar1->foo+x->bar2 if you include returns.
A callchain is a pure stack, a return pops the top most entry, the above program can only give 3 possible callchains:
a) foo b) foo, bar1 c) foo, bar2
Furthermore, the LBR is about any branch, callchains are about function calls.
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