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From: Peter Zijlstra <peterz@infradead.org>
To: Michel Lespinasse <michel@lespinasse.org>
Cc: Li RongQing <lirongqing@baidu.com>,
	dbueso@suse.de, mingo@kernel.org, linux-kernel@vger.kernel.org
Subject: Re: [PATCH] rbtree: stop iteration early in rb_find_first
Date: Wed, 25 Aug 2021 14:29:19 +0200	[thread overview]
Message-ID: <YSY3nwCY/IDl1hpj@hirez.programming.kicks-ass.net> (raw)
In-Reply-To: <20210825115332.GA4645@lespinasse.org>

On Wed, Aug 25, 2021 at 04:53:32AM -0700, Michel Lespinasse wrote:
> On Wed, Aug 25, 2021 at 01:39:26PM +0200, Peter Zijlstra wrote:
> > On Wed, Aug 25, 2021 at 05:59:48PM +0800, Li RongQing wrote:
> > > stop iteration if match is not NULL and result of cmp is
> > > not zero, this means the matched node has been found, and
> > > the node with same key has been passed
> > > 
> > > Signed-off-by: Li RongQing <lirongqing@baidu.com>
> > > ---
> > >  include/linux/rbtree.h | 3 +++
> > >  1 file changed, 3 insertions(+)
> > > 
> > > diff --git a/include/linux/rbtree.h b/include/linux/rbtree.h
> > > index d31ecaf4fdd3..2689771df9bb 100644
> > > --- a/include/linux/rbtree.h
> > > +++ b/include/linux/rbtree.h
> > > @@ -324,6 +324,9 @@ rb_find_first(const void *key, const struct rb_root *tree,
> > >  		} else if (c > 0) {
> > >  			node = node->rb_right;
> > >  		}
> > > +
> > > +		if (match && c)
> > > +			break;
> > >  	}
> > >  
> > >  	return match;
> > 
> > Acked-by: Peter Zijlstra (Intel) <peterz@infradead.org>
> 
> NAK. This looked slightly wrong before, and is more wrong after.
> 
> Before:
> there was this weird condition  if (c <= 0) {} else if (c > 0) {} ,
> making you wonder what the third possibility may be. Easy fix would be
> to remove the second condition.
> 
> After:
> say the key is equal the root, so the code sets match=root and goes left.
> Then it stops searching because match is set and c<0.
> This doesn't work, the code needs to keep searching for the leftmost match.

I'm not following you. If c!=0 the key didn't match. If c<0 the key is
less than the one we're looking for, meaning we've already found the
left-most matching key, idem for c>0.

More specifically, can you draw me a (binary) tree with elements: A B B
B C, such that a search for B might have match set, hit c!=1 and not
have found the leftmost B ?

  parent reply	other threads:[~2021-08-25 12:30 UTC|newest]

Thread overview: 13+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2021-08-25  9:59 Li RongQing
2021-08-25 11:39 ` Peter Zijlstra
2021-08-25 11:53   ` Michel Lespinasse
2021-08-25 11:58     ` Michel Lespinasse
2021-08-25 12:31       ` Peter Zijlstra
2021-08-25 13:21       ` Peter Zijlstra
2021-08-25 16:08         ` 答复: " Li,Rongqing
     [not found]         ` <90ea3457ddc7485fbc8db5f7ca5b07ab@baidu.com>
2021-08-25 17:18           ` Peter Zijlstra
2021-08-25 18:26             ` 答复: " Li,Rongqing
2021-08-25 19:20               ` Peter Zijlstra
2021-08-26  5:03                 ` 答复: " Li,Rongqing
2021-08-25 12:29     ` Peter Zijlstra [this message]
2021-08-25 12:32       ` Peter Zijlstra

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