On Fri, 30 Nov 2012 10:52:15 -0300, Arnaldo Carvalho de Melo wrote:
> Em Fri, Nov 30, 2012 at 02:29:43PM +0100, Jiri Olsa escreveu:
>> On Thu, Nov 29, 2012 at 03:38:40PM +0900, Namhyung Kim wrote:
>> > +#define __HPP_COLOR_PERCENT_FN(_type, _field)
>> > \
>> > +static int hpp__color_##_type(struct perf_hpp *hpp, struct hist_entry
>> > *he) \
>> > +{
>> > \
>> > + int ret;
>> > \
>> > + double percent = 0.0;
>> > \
>> > + struct hists *hists = he->hists;
>> > \
>> > +
>> > \
>> > + if (hists->stats.total_period)
>> > \
>> > + percent = 100.0 * he->stat._field / hists->stats.total_period;
>> > \
>> > +
>> > \
>> > + ret = percent_color_snprintf(hpp->buf, hpp->size, " %6.2f%%", percent);
>> > \
>> > +
>> > \
>> > + if (symbol_conf.event_group) {
>> > \
>> > + int i;
>> > \
>> > + struct perf_evsel *evsel = hists_to_evsel(hists);
>> > \
>> > + struct hist_entry *pair;
>> > \
>> > + int nr_members = evsel->nr_members;
>> > \
>> > + double *percents;
>> > \
>> > +
>> > \
>> > + if (nr_members <= 1)
>> > \
>> > + return ret;
>> > \
>> > +
>> > \
>> > + percents = zalloc(sizeof(*percents) * nr_members);
>> > \
>> > + if (percents == NULL) {
>> > \
>> > + pr_warning("Not enough memory!\n");
>> > \
>> > + return ret;
>> > \
>> > + }
>> > \
>
> Why do we need to zalloc this percents array?
>
>> > + list_for_each_entry(pair, &he->pairs.head, pairs.node) {
>> > \
>> > + u64 period = pair->stat._field;
>> > \
>> > + u64 total = pair->hists->stats.total_period;
>> > \
>> > +
>> > \
>> > + if (!total)
>> > \
>> > + continue;
>> > \
>> > +
>> > \
>> > + evsel = hists_to_evsel(pair->hists);
>> > \
>> > + i = perf_evsel__group_idx(evsel);
>> > \
>> > + percents[i] = 100.0 * period / total;
>> > \
>
> Why not use percent_color_snprintf here, using some "%*s" thing that
> uses i to position the output in the right column? This way the
> following loop can be ditched as well. No?
It's because it's possible that an entry didn't have pairs for every
group member. Say, there's a group consists of 3 members (1 leader + 2
member). It's quite legitimate that a leader hist entry has just one
pair for a member and miss another. So I allocated a zero-filled array,
and filled what it has, and print them all in the for loop below.
Thanks,
Namhyung
>
>> > + }
>> > \
>> > +
>> > \
>> > + for (i = 1; i < nr_members; i++) {
>> > \
>> > + ret += percent_color_snprintf(hpp->buf + ret,
>> > \
>> > + hpp->size - ret,
>> > \
>> > + " %6.2f%%", percents[i]);
>> > \
>> > + }
>> > \
>> > + free(percents);
>> > \
>> > + }
>> > \
>> > + return ret;
>> > \
>
> - Arnaldo
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