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From:   Wei Yang <richard.weiyang@linux.alibaba.com>
To:     rostedt@goodmis.org, mingo@redhat.com
Cc:     linux-kernel@vger.kernel.org,
        Wei Yang <richard.weiyang@linux.alibaba.com>
Subject: [PATCH 4/6] ftrace: simplify the calculation of page number for ftrace_page->records
Date:   Mon, 31 Aug 2020 11:11:02 +0800
Message-Id: <20200831031104.23322-5-richard.weiyang@linux.alibaba.com>
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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.

Signed-off-by: Wei Yang <richard.weiyang@linux.alibaba.com>
---
 kernel/trace/ftrace.c | 7 ++++---
 1 file changed, 4 insertions(+), 3 deletions(-)

diff --git a/kernel/trace/ftrace.c b/kernel/trace/ftrace.c
index 9021e16fa079..15fcfa16895d 100644
--- a/kernel/trace/ftrace.c
+++ b/kernel/trace/ftrace.c
@@ -3127,18 +3127,19 @@ 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:
-- 
2.20.1 (Apple Git-117)