From mboxrd@z Thu Jan  1 00:00:00 1970
From: Theodore Ts'o <tytso@mit.edu>
Subject: Re: [PATCH v2 1/5] ext4: improve ext4_es_can_be_merged() to avoid a
 potential overflow
Date: Sun, 10 Mar 2013 20:43:58 -0400
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Cc: linux-ext4@vger.kernel.org, Zheng Liu <wenqing.lz@taobao.com>,
	Dmitry Monakhov <dmonakhov@openvz.org>
To: Zheng Liu <gnehzuil.liu@gmail.com>
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On Wed, Mar 06, 2013 at 10:17:11PM +0800, Zheng Liu wrote:
> +	if (ext4_es_status(es1) ^ ext4_es_status(es2))
>  		return 0;
>  
> -	if (ext4_es_status(es1) != ext4_es_status(es2))

Did you have a reason why changed != to ^?  

It's identical from a functional perspective, but it's less obvious to
future readers of the code what's going on.  I tried checking to see
if GCC did any better optimizing the code, but it doesn't seem to make
any difference.  I'm going to switch it back to !=....

> +	/* we need to check delayed extent is without unwritten status */
> +	if (ext4_es_is_delayed(es1) && !ext4_es_is_unwritten(es1))
> +		return 1;

I'm not sure why we need to check the unwritten status?  Under what
circumstances would we have an extent marked as under delayed
allocation but also unwritten?

							- Ted

This is how I've restructured this function for now mainly to make it
easier to understand;

static int ext4_es_can_be_merged(struct extent_status *es1,
				 struct extent_status *es2)
{
	if (ext4_es_status(es1) != ext4_es_status(es2))
		return 0;

	if (((__u64) es1->es_len) + es2->es_len > 0xFFFFFFFFULL)
		return 0;

	if (((__u64) es1->es_lblk) + es1->es_len != es2->es_lblk)
		return 0;

	if ((ext4_es_is_written(es1) || ext4_es_is_unwritten(es1)) &&
	    (ext4_es_pblock(es1) + es1->es_len == ext4_es_pblock(es2)))
		return 1;

	if (ext4_es_is_hole(es1))
		return 1;

	/* we need to check delayed extent is without unwritten status */
	if (ext4_es_is_delayed(es1) && !ext4_es_is_unwritten(es1))
		return 1;

	return 0;
}