[Ocfs2-devel] [PATCH 3/3] ocfs2/xattr: Proper hash collision handle in bucket division.v2

[Joel Becker]

On Fri, Oct 24, 2008 at 12:43:53AM -0700, Joel Becker wrote:
> On Fri, Oct 24, 2008 at 02:15:35PM +0800, Tao Ma wrote:
> > Joel Becker wrote:
> >> 	I have an idea to do what you're doing, but cleaner.  I'll look
> >> at cmp_xe and try again in the morning.
> > cool, so waiting for your perfect code. ;)
> 
> Zing!  You got me :-)

	How's this (untested)?

/*
 * If this ocfs2_xattr_header covers more than one hash value, find a
 * place where the hash value changes.  Try to find the most even split.
 * The most common case is that all entries have different hash values,
 * and the first check we make will find a place to split.
 */
static int ocfs2_xattr_find_divide_pos(struct ocfs2_xattr_header *xh)
{
	struct ocfs2_xattr_entry *entries = &xh->xh_entries;
	int count = le16_to_cpu(xh->xh_count);
	int delta, middle = count / 2;

	/*  
	 * We start at the middle.  Each step gets farther away in both
	 * directions.  We therefore hit the change in hash value
	 * nearest to the middle.  Note that this loop does not execute for
	 * count < 2.
	 */
	for (delta = 0; delta < middle; delta++) {
		/* Let's check delta earlier than middle */
		if (!cmp_ex(&entries[middle - delta - 1], 
			    &entries[middle - delta]))
			return middle - delta;
		/* For even counts, don't walk off the end */
		if ((middle + delta + 1) == count)
			continue;
		/* Now try delta past middle */
		if (!cmp_ex(&entries[middle + delta],
			    &entries[middle + delta + 1]))
			return middle + delta + 1;
	}

	/* Every entry had the same hash */
	return count;
}

Joel

-- 

"Reader, suppose you were and idiot.  And suppose you were a member of
 Congress.  But I repeat myself."
	- Mark Twain

Joel Becker
Principal Software Developer
Oracle
E-mail: joel.becker at oracle.com
Phone: (650) 506-8127