Check whether denominator expression x * (x - 1) * 1000 mod {2^32, 2^64}
produce zero and skip stddev computation in that case.
For now don't care about rec->counter * rec->counter overflow because
rec->time * rec->time overflow will likely happen earlier.
Cc: stable@xxxxxxxxxxxxxxx
Fixes: e31f7939c1c27 ("ftrace: Avoid potential division by zero in function profiler")
Signed-off-by: Nikolay Kuratov <kniv@xxxxxxxxxxxxxx>
---
v2: Instead of splitting the division into two store denominator into
separate variable and zero-check it
kernel/trace/ftrace.c | 27 ++++++++++++---------------
1 file changed, 12 insertions(+), 15 deletions(-)
diff --git a/kernel/trace/ftrace.c b/kernel/trace/ftrace.c
index 728ecda6e8d4..9ea6609588cf 100644
--- a/kernel/trace/ftrace.c
+++ b/kernel/trace/ftrace.c
@@ -540,6 +540,7 @@ static int function_stat_show(struct seq_file *m, void *v)
static struct trace_seq s;
unsigned long long avg;
unsigned long long stddev;
+ unsigned long long stddev_denom;
#endif
guard(mutex)(&ftrace_profile_lock);
@@ -559,23 +560,19 @@ static int function_stat_show(struct seq_file *m, void *v)
#ifdef CONFIG_FUNCTION_GRAPH_TRACER
seq_puts(m, " ");
- /* Sample standard deviation (s^2) */
- if (rec->counter <= 1)
- stddev = 0;
- else {
- /*
- * Apply Welford's method:
- * s^2 = 1 / (n * (n-1)) * (n * \Sum (x_i)^2 - (\Sum x_i)^2)
- */
+ /*
+ * Variance formula:
+ * s^2 = 1 / (n * (n-1)) * (n * \Sum (x_i)^2 - (\Sum x_i)^2)
+ * Maybe Welford's method is better here?
+ * Divide only by 1000 for ns^2 -> us^2 conversion.
+ * trace_print_graph_duration will divide by 1000 again.
+ */
+ stddev = 0;
+ stddev_denom = rec->counter * (rec->counter - 1) * 1000;
+ if (stddev_denom) {
stddev = rec->counter * rec->time_squared -
rec->time * rec->time;
-
- /*
- * Divide only 1000 for ns^2 -> us^2 conversion.
- * trace_print_graph_duration will divide 1000 again.
- */
- stddev = div64_ul(stddev,
- rec->counter * (rec->counter - 1) * 1000);
+ stddev = div64_ul(stddev, stddev_denom);
}
trace_seq_init(&s);
--
2.34.1
