ftrace: fix conversion of task state to char in latency tracer

The conversion of task states to a character in the sched_switch tracer (part
of latency tracer infrastructure), seems to be incorrect. We currently do it
by indexing into the state_to_char array using the state value. The state
values do not map directly into the array index and are thus incorrect. The
following patch addresses this issue. This is also what is being done even
in the show_task() routine in kernel/sched.c

The patch has been compile and run tested.

Signed-off-by: Ankita Garg <ankita@in.ibm.com>
Signed-off-by: Ingo Molnar <mingo@elte.hu>
Signed-off-by: Thomas Gleixner <tglx@linutronix.de>

diff --git a/kernel/trace/trace.c b/kernel/trace/trace.c
index 0eef050..9197782 100644 (file)
--- a/kernel/trace/trace.c
+++ b/kernel/trace/trace.c
@@ -1208,6 +1208,7 @@ print_lat_fmt(struct trace_iterator *iter, unsigned int trace_idx, int cpu)
        char *comm;
        int S, T;
        int i;
+       unsigned state;
 
        if (!next_entry)
                next_entry = entry;
@@ -1238,11 +1239,11 @@ print_lat_fmt(struct trace_iterator *iter, unsigned int trace_idx, int cpu)
                break;
        case TRACE_CTX:
        case TRACE_WAKE:
-               S = entry->ctx.prev_state < sizeof(state_to_char) ?
-                       state_to_char[entry->ctx.prev_state] : 'X';
                T = entry->ctx.next_state < sizeof(state_to_char) ?
                        state_to_char[entry->ctx.next_state] : 'X';
 
+               state = entry->ctx.prev_state ? __ffs(entry->ctx.prev_state) + 1 : 0;
+               S = state < sizeof(state_to_char) - 1 ? state_to_char[state] : 'X';
                comm = trace_find_cmdline(entry->ctx.next_pid);
                trace_seq_printf(s, " %5d:%3d:%c %s %5d:%3d:%c %s\n",
                                 entry->ctx.prev_pid,