From mboxrd@z Thu Jan  1 00:00:00 1970
From: Arun Sharma <asharma@fb.com>
Subject: Re: [PATCH 0/2] perf: add sort by inclusive time functionality (v2)
Date: Mon, 12 Mar 2012 11:05:16 -0700
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References: <1331160079-13821-1-git-send-email-asharma@fb.com> <20120308072901.GC20784@elte.hu> <20120308153130.GC7976@somewhere.redhat.com> <4F58FF47.6090504@fb.com> <4F5DA294.6070700@lge.com>
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To: Namhyung Kim <namhyung.kim@lge.com>
Cc: Frederic Weisbecker <fweisbec@gmail.com>, Ingo Molnar <mingo@elte.hu>, linux-kernel@vger.kernel.org, Arnaldo Carvalho de Melo <acme@redhat.com>, Mike Galbraith <efault@gmx.de>, Paul Mackerras <paulus@samba.org>, Peter Zijlstra <peterz@infradead.org>, Stephane Eranian <eranian@google.com>, Tom Zanussi <tzanussi@gmail.com>, linux-perf-users@vger.kernel.org

On 3/12/12 12:15 AM, Namhyung Kim wrote:

>
> I think it's because of the shared hist_entry. If a callchain is a
> subset of another, it will be marked as inclusive so that it cannot be
> contributed to total period. Say, there're two chains - X (a -> b -> c)
> and Y (a -> b), once __hists__add_entry_inclusive() was called on X, we
> have:
>
> a -> b -> c
> a -> b (inclusive)
> a (inclusive)
>
> And then, calling the function on Y should make:
>
> a -> b
> a (inclusive)
>
> However, since both callchains are in tree already they'll be shared and
> marked *inclusive*. Thus the total period will not increased at all for
> Y. Also I guess the reverse case - add Y first, and then X - will have
> the same result.

Thanks for figuring this out. Looks like using a single bit 
(he->inclusive) is insufficient. How about:

struct hist_entry {
	u64                     period;
         u64                     period_self;
	..
};

Normal mode: period_self == period.
Inclusive mode: period_self will be zero for inclusive hist_entries.
Shared entries: we sum up both period and period_self.

We can then compute total_period by summing up period_self.

  -Arun