linux-kernel - Re: 答复: 答复: [PATCH] rbtree: stop iteration early in rb_find

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Message-ID: <20210825192056.GF17784@worktop.programming.kicks-ass.net>
Date:   Wed, 25 Aug 2021 21:20:56 +0200
From:   Peter Zijlstra <peterz@...radead.org>
To:     "Li,Rongqing" <lirongqing@...du.com>
Cc:     Michel Lespinasse <michel@...pinasse.org>,
        "dbueso@...e.de" <dbueso@...e.de>,
        "mingo@...nel.org" <mingo@...nel.org>,
        "linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>
Subject: Re: 答复: 答复: [PATCH] rbtree:
 stop iteration early in rb_find_first

On Wed, Aug 25, 2021 at 06:26:59PM +0000, Li,Rongqing wrote:
> 
>                        10
>                      /
>                     5
>                        \
>                       10
> 
> the above case should not exist. like below, when second 10 is inserted, it should be inserted to right leaf
>                        10
>                      /
>                     5
> 
> as a result, it should be
> 
>                        10
>                      /     \
>                     5      10
> 
> since 10 is not less 10, so new 10 is inserted to right.

You're right that rb_add() does a tail-add for elements it considers
equal -- there is actually code in the tree that relies on this.

But that doesn't mean rb_find_first() should go right, it *must*not*,
because then it wouldn't find the 'first' aka 'leftmost' instance of the
equal elements.

Also, you're only considering building the tree in-order with rb_add(),
trees get modified all the time and the pattern Michel gave is perfectly
valid (also see rb_prev()).

Sure the snippet is not a balanced tree, but you can construct the
pattern as part of a larger tree just fine, just add some elements:

	 10(b)
	/  \
       5   10(c)
        \
	10(a)

is a tree that is balanced (remember, RB trees only require the left and
right depths to no more than double -- as opposed to AVL trees, which
have a tighter constraint). This tree has order: 5, 10(a), 10(b), 10(c).
Also note that the tree rotations are stable -- they must be since they
do not refence the order function.

As such, if we rb_find_first() for 10, we must find 10(a), the leftmost
10 in the tree.