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>

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: