ftrace: Simplify the calculation of page number for ftrace_page->records
Based on the following two reasones, we could simplify the calculation:
- If the number after roundup count is not power of 2, we would
definitely have more than 1 empty page with a higher order.
- get_count_order() just return current order, so one lower order
could meet the requirement.
The calculation could be simplified by lower one order level when pages
are not power of 2.
Link: https://lkml.kernel.org/r/20200831031104.23322-5-richard.weiyang@linux.alibaba.com
Signed-off-by: Wei Yang <richard.weiyang@linux.alibaba.com>
Signed-off-by: Steven Rostedt (VMware) <rostedt@goodmis.org>
diff --git a/kernel/trace/ftrace.c b/kernel/trace/ftrace.c
index c51a91a..c3be18b 100644
--- a/kernel/trace/ftrace.c
+++ b/kernel/trace/ftrace.c
@@ -3129,18 +3129,19 @@ static int ftrace_update_code(struct module *mod, struct ftrace_page *new_pgs)
static int ftrace_allocate_records(struct ftrace_page *pg, int count)
{
int order;
- int cnt;
+ int pages, cnt;
if (WARN_ON(!count))
return -EINVAL;
- order = get_count_order(DIV_ROUND_UP(count, ENTRIES_PER_PAGE));
+ pages = DIV_ROUND_UP(count, ENTRIES_PER_PAGE);
+ order = get_count_order(pages);
/*
* We want to fill as much as possible. No more than a page
* may be empty.
*/
- while ((PAGE_SIZE << order) / ENTRY_SIZE >= count + ENTRIES_PER_PAGE)
+ if (!is_power_of_2(pages))
order--;
again:
