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Date:	Wed, 14 Mar 2012 12:48:04 -0400
From:	Ted Ts'o <tytso@....edu>
To:	Zach Brown <zab@...bo.net>
Cc:	Yongqiang Yang <xiaoqiangnk@...il.com>,
	Phillip Susi <phillsusi@...il.com>,
	Andreas Dilger <adilger@...mcloud.com>,
	Lukas Czerner <lczerner@...hat.com>,
	Jacek Luczak <difrost.kernel@...il.com>,
	"linux-ext4@...r.kernel.org" <linux-ext4@...r.kernel.org>,
	linux-fsdevel <linux-fsdevel@...r.kernel.org>,
	LKML <linux-kernel@...r.kernel.org>,
	"linux-btrfs@...r.kernel.org" <linux-btrfs@...r.kernel.org>
Subject: Re: getdents - ext4 vs btrfs performance

On Wed, Mar 14, 2012 at 10:17:37AM -0400, Zach Brown wrote:
> 
> >We could do this if we have two b-trees, one indexed by filename and
> >one indexed by inode number, which is what JFS (and I believe btrfs)
> >does.
> 
> Typically the inode number of the destination inode isn't used to index
> entries for a readdir tree because of (wait for it) hard links.  You end
> up right back where you started with multiple entries per key.

Well, if you are using 32-bit (or even 48-bit) inode numbers and a
64-bit telldir cookie, it's possible to make the right thing happen.
But yes, if you are using 32-bit inode numbers and a 32-bit telldir
cookie, dealing with what happens when you have multiple hard links to
the same inode in the same directory gets tricky.

> A painful solution is to have the key in the readdir tree allocated by
> the tree itself -- count key populations in subtrees per child pointer
> and use that to find free keys.

One thing that might work is to have a 16-bit extra field in the
directory entry that gives an signed offset to the inode number so
that such that inode+offset is a unique value within the btree sorted
by inode+offset number.  Since this tree is only used for returning
entries in an optimal (or as close to optimal as can be arranged)
order, we could get away with that.

				- Ted
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