perf evlist: Introduce set_filter_pids method

We need to filter multiple pids in trace, i.e. trace itself,
gnome-terminal, X.org, etc.

Cc: Adrian Hunter <adrian.hunter@intel.com>
Cc: Borislav Petkov <bp@suse.de>
Cc: David Ahern <dsahern@gmail.com>
Cc: Don Zickus <dzickus@redhat.com>
Cc: Frederic Weisbecker <fweisbec@gmail.com>
Cc: Jiri Olsa <jolsa@redhat.com>
Cc: Mike Galbraith <efault@gmx.de>
Cc: Namhyung Kim <namhyung@kernel.org>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Stephane Eranian <eranian@google.com>
Link: http://lkml.kernel.org/n/tip-frtpkg7qapqwf7asa35wf8am@git.kernel.orgSigned-off-by: NArnaldo Carvalho de Melo <acme@redhat.com>
上级 241b057c
......@@ -1085,19 +1085,38 @@ int perf_evlist__set_filter(struct perf_evlist *evlist, const char *filter)
return err;
}
int perf_evlist__set_filter_pid(struct perf_evlist *evlist, pid_t pid)
int perf_evlist__set_filter_pids(struct perf_evlist *evlist, size_t npids, pid_t *pids)
{
char *filter;
int ret;
int ret = -1;
size_t i;
if (asprintf(&filter, "common_pid != %d", pid) < 0)
return -1;
for (i = 0; i < npids; ++i) {
if (i == 0) {
if (asprintf(&filter, "common_pid != %d", pids[i]) < 0)
return -1;
} else {
char *tmp;
if (asprintf(&tmp, "%s && common_pid != %d", filter, pids[i]) < 0)
goto out_free;
free(filter);
filter = tmp;
}
}
ret = perf_evlist__set_filter(evlist, filter);
out_free:
free(filter);
return ret;
}
int perf_evlist__set_filter_pid(struct perf_evlist *evlist, pid_t pid)
{
return perf_evlist__set_filter_pids(evlist, 1, &pid);
}
bool perf_evlist__valid_sample_type(struct perf_evlist *evlist)
{
struct perf_evsel *pos;
......
......@@ -78,6 +78,7 @@ int perf_evlist__add_newtp(struct perf_evlist *evlist,
int perf_evlist__set_filter(struct perf_evlist *evlist, const char *filter);
int perf_evlist__set_filter_pid(struct perf_evlist *evlist, pid_t pid);
int perf_evlist__set_filter_pids(struct perf_evlist *evlist, size_t npids, pid_t *pids);
struct perf_evsel *
perf_evlist__find_tracepoint_by_id(struct perf_evlist *evlist, int id);
......
Markdown is supported
0% .
You are about to add 0 people to the discussion. Proceed with caution.
先完成此消息的编辑!
想要评论请 注册