Messages in this thread | | | From | Milian Wolff <> | Subject | Re: [PATCH v6 1/6] perf report: properly handle branch count in match_chain | Date | Fri, 20 Oct 2017 13:38:23 +0200 |
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On Freitag, 20. Oktober 2017 12:21:35 CEST Milian Wolff wrote: > On Donnerstag, 19. Oktober 2017 17:01:08 CEST Namhyung Kim wrote: > > Hi Andi, > > > > On Thu, Oct 19, 2017 at 06:55:19AM -0700, Andi Kleen wrote: > > > On Thu, Oct 19, 2017 at 12:59:14PM +0200, Milian Wolff wrote: > > > > On Donnerstag, 19. Oktober 2017 00:41:04 CEST Andi Kleen wrote: > > > > > Milian Wolff <milian.wolff@kdab.com> writes: > > > > > > +static enum match_result match_address_dso(struct dso *left_dso, > > > > > > u64 > > > > > > left_ip, + struct dso *right_dso, u64 right_ip) > > > > > > +{ > > > > > > + if (left_dso == right_dso && left_ip == right_ip) > > > > > > + return MATCH_EQ; > > > > > > + else if (left_ip < right_ip) > > > > > > + return MATCH_LT; > > > > > > + else > > > > > > + return MATCH_GT; > > > > > > +} > > > > > > > > > > So why does only the first case check the dso? Does it not matter > > > > > for the others? > > > > > > > > > > Either should be checked by none or by all. > > > > > > > > I don't see why it should be checked. It is only required to prevent > > > > two > > > > addresses to be considered equal while they are not. So only the one > > > > check is required, otherwise we return either LT or GT. > > > > > > When the comparison is always in the same process (which I think > > > is not the case) just checking the addresses is sufficient. If they are > > > not then you always need to check the DSO and only compare inside the > > > same DSO. > > > > As far as I know, the node->ip is a relative address (inside a DSO). > > So it should compare the dso as well even in the same process. > > Sorry guys, I seem to be slow at understanding your review comments. > > match_address_dso should impose a sort order on two relative addresses. The > order should ensure that relative addresses in a different DSO are not > considered equal. But if the DSOs are different, it doesn't matter whether > we return LT or GT - or? > > Put differently, how would you write this function to take care of the DSO > in the other two branches? I.e. what to return if the DSOs are different - > a MATCH_ERROR?
Thinking a bit more about this. Are you guys maybe hinting at my implementation breaking the strict ordering rules (is that the right word?). I.e. a < b && b > a iff a == b ? Potentially my implementation would break this assumption when the relative IPs are the same, but the DSO is different.
So is this what you want:
+static enum match_result match_address_dso(struct dso *left_dso, u64 left_ip, + struct dso *right_dso, u64 right_ip) +{ + if (left_dso == right_dso && left_ip == right_ip) + return MATCH_EQ; + else if (left_dso < right_dso || left_ip < right_ip) + return MATCH_LT; + else + return MATCH_GT; +}
Thanks
-- Milian Wolff | milian.wolff@kdab.com | Senior Software Engineer KDAB (Deutschland) GmbH&Co KG, a KDAB Group company Tel: +49-30-521325470 KDAB - The Qt Experts
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