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SubjectRe: [PATCH 13/23] perf hists: Support filtering in hierarchy mode
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Hi Jiri,

2016-02-10 (수), 12:51 +0100, Jiri Olsa:
> On Fri, Feb 05, 2016 at 10:01:45PM +0900, Namhyung Kim wrote:
>
> SNIP
>
> > + parent = parent->parent_he;
> > + }
> > + }
> > +
> >   if (h->filtered)
> >   return;
> >  
> > @@ -1592,28 +1613,122 @@ static void hists__filter_by_type(struct hists *hists, int type, filter_fn_t fil
> >   }
> >  }
> >  
> > +static void hists__filter_hierarchy(struct hists *hists, int type, const void *arg)
> > +{
> > + struct rb_node *nd;
> > + struct rb_root tmp = RB_ROOT;
> > + bool saved_unfolded;
> > +
> > + hists->stats.nr_non_filtered_samples = 0;
> > +
> > + hists__reset_filter_stats(hists);
> > + hists__reset_col_len(hists);
> > +
> > + nd = rb_first(&hists->entries);
> > + while (nd) {
> > + struct hist_entry *h = rb_entry(nd, struct hist_entry, rb_node);
> > + int ret;
> > +
> > + ret = hist_entry__filter(h, type, arg);
> > +
> > + /*
> > +  * case 1. non-matching type
> > +  * zero out the period, set filtered and move to child
> > +  */
> > + if (ret < 0) {
> > + memset(&h->stat, 0, sizeof(h->stat));
> > + h->filtered |= (1 << type);
> > +
> > + /* force to go down in the hierarchy */
> > + saved_unfolded = h->unfolded;
> > + h->unfolded = true;
> > +
> > + nd = rb_hierarchy_next(&h->rb_node);
> > + h->unfolded = saved_unfolded;
> > + }
> > + /*
> > +  * case 2. matched (filter out)
> > +  * set filtered and move to next
> > +  */
> > + else if (ret == 1) {
>
> slightly confused in here.. so ret == 1 means we matched
> but for example hist_entry__sym_filter:
>
> +       return sym && (!he->ms.sym || !strstr(he->ms.sym->name, sym));
>
> returns 1 if we DONT match the symbol

Yes, this was confusing.  I meant it's a matching *type* but
the filter is not matched.


>
> I always got confused by non/filtered stuff.. I think we should change
> the those names, but haven't come with anything so far..

Yeah, me too.. Filter can have both semantics - opt-in and opt-out.
It also made me hard to choose a good name.

Thanks,
Namhyung

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