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Message-ID: <3yrtyxb6mwtwrov4vngtcy34pl77easph6hueo5m3nxlqx6o5c@f4frvl5cxees>
Date: Thu, 26 Jun 2025 15:58:03 -0400
From: "Liam R. Howlett" <Liam.Howlett@...cle.com>
To: Dev Jain <dev.jain@....com>
Cc: akpm@...ux-foundation.org, richard.weiyang@...il.com,
maple-tree@...ts.infradead.org, linux-mm@...ck.org,
linux-kernel@...r.kernel.org
Subject: Re: [PATCH 1/2] maple tree: Clean up mtree_range_walk()
* Dev Jain <dev.jain@....com> [250626 13:19]:
> The special casing for offset == 0 is being done because min will stay
> mas->min in this case. So refactor the code to use the while loop for
> setting the max and getting the corresponding offset, and only set the
> min for offset > 0.
>
> Signed-off-by: Dev Jain <dev.jain@....com>
> ---
> lib/maple_tree.c | 11 +++--------
> 1 file changed, 3 insertions(+), 8 deletions(-)
>
> diff --git a/lib/maple_tree.c b/lib/maple_tree.c
> index 0e85e92c5375..6c89e6790fb5 100644
> --- a/lib/maple_tree.c
> +++ b/lib/maple_tree.c
> @@ -2770,13 +2770,8 @@ static inline void *mtree_range_walk(struct ma_state *mas)
> end = ma_data_end(node, type, pivots, max);
> prev_min = min;
> prev_max = max;
> - if (pivots[0] >= mas->index) {
> - offset = 0;
> - max = pivots[0];
> - goto next;
> - }
>
This new line should be dropped.
> - offset = 1;
> + offset = 0;
> while (offset < end) {
This should now be a do {} while();
> if (pivots[offset] >= mas->index) {
> max = pivots[offset];
> @@ -2784,9 +2779,9 @@ static inline void *mtree_range_walk(struct ma_state *mas)
> }
> offset++;
> }
There should be a new line here.
> + if (likely(offset))
> + min = pivots[offset - 1] + 1;
>
> - min = pivots[offset - 1] + 1;
> -next:
> slots = ma_slots(node, type);
> next = mt_slot(mas->tree, slots, offset);
> if (unlikely(ma_dead_node(node)))
> --
> 2.30.2
>
The current way will check pivot 0, then skip the main loop. Pivot 0
has an equal chance of being the range you are looking for, but that
probability increases based on a lower number of entries in the node.
The root node, which we always pass through, can have as little as two
entries, so then it's 50/50 you want pivot 0.
With the way you've written it, it will check offset < end (which will
always be the case for the first time), then do the same work as the
out-of-loop check, then check offset an extra time after the loop.
If it's written with a do {} while, the initial check of offset < end is
avoided, but you will end up checking offset an extra time to see if min
needs to be set.
If pivot 0 is NOT the entry you want, then the current code will check
pivot 0, then execute the loop one less time. In the even of a root
node with 2 entries, it will not enter the loop at all.
So, the way it is written is less code execution by avoiding unnecessary
assignment of min and checks (of offset == 0 and offset < end).
Thanks,
Liam
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