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: [email protected] Fixes: e31f7939c1c27 ("ftrace: Avoid potential division by zero in function profiler") Signed-off-by: Nikolay Kuratov <[email protected]> --- 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
