On 7/27/16 6:54 AM, Tomoki Sekiyama wrote:
sched_out_state() converts the prev_state u64 bitmask to a char in
a wrong way, which may cause wrong results of 'perf sched latency'.
This patch fixes the conversion.
Signed-off-by: Tomoki Sekiyama <tomoki.sekiyama...@hitachi.com>
Cc: Jiri Olsa <jo...@kernel.org>
Cc: David Ahern <dsah...@gmail.com>
Cc: Namhyung Kim <namhy...@kernel.org>
Cc: Peter Zijlstra <a.p.zijls...@chello.nl>
Cc: Masami Hiramatsu <mhira...@kernel.org>
---
tools/perf/builtin-sched.c | 3 ++-
1 file changed, 2 insertions(+), 1 deletion(-)
diff --git a/tools/perf/builtin-sched.c b/tools/perf/builtin-sched.c
index 0dfe8df..eb2f7f4 100644
--- a/tools/perf/builtin-sched.c
+++ b/tools/perf/builtin-sched.c
@@ -71,6 +71,7 @@ struct sched_atom {
};
#define TASK_STATE_TO_CHAR_STR "RSDTtZXxKWP"
+#define TASK_STATE_MASK 0x7ff
The mask should not be needed and looking at top of tree there are 2 new
states (N and n) that need to be added.
enum thread_state {
THREAD_SLEEPING = 0,
@@ -899,7 +900,7 @@ static char sched_out_state(u64 prev_state)
{
const char *str = TASK_STATE_TO_CHAR_STR;
- return str[prev_state];
+ return str[ffs(prev_state & TASK_STATE_MASK)];
}
static int
Handle unknown bits with '?' like the kernel does (see task_state_char
and sched_show_task).
