linux-kernel - Re: [PATCH] aoe: error printed 1 too early

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Message-ID: <20090225170901.GA26634@coraid.com>
Date:	Wed, 25 Feb 2009 12:09:02 -0500
From:	Ed Cashin <ecashin@...aid.com>
To:	Roel Kluin <roel.kluin@...il.com>
Cc:	lkml <linux-kernel@...r.kernel.org>,
	Andrew Morton <akpm@...ux-foundation.org>
Subject: Re: [PATCH] aoe: error printed 1 too early

On Wed, Feb 25, 2009 at 02:38:54PM +0100, Roel Kluin wrote:
> with while (i-- > 0); i reaches -1 after the loop, so the test below is printed
> one too early: 0 still means success.
> 
> Signed-off-by: Roel Kluin <roel.kluin@...il.com>
> ---
> diff --git a/drivers/block/aoe/aoedev.c b/drivers/block/aoe/aoedev.c
> index cc25057..eeea477 100644
> --- a/drivers/block/aoe/aoedev.c
> +++ b/drivers/block/aoe/aoedev.c
> @@ -173,7 +173,7 @@ skbfree(struct sk_buff *skb)
>  		return;
>  	while (atomic_read(&skb_shinfo(skb)->dataref) != 1 && i-- > 0)
>  		msleep(Sms);
> -	if (i <= 0) {
> +	if (i < 0) {
>  		printk(KERN_ERR
>  			"aoe: %s holds ref: %s\n",
>  			skb->dev ? skb->dev->name : "netif",
> 

You are talking about the case where, when entering the loop on the
"last chance" iteration, the states are,

  atomic_read(&skb_shinfo(skb)->dataref == 1
  i == 0

... in which case, the skb can and should be used but is not used
because after the value of i is tested it becomes -1.

Right?

Sounds like a good catch.  Did you find it because you saw the "cannot
free skb" message?

-- 
  Ed Cashin <ecashin@...aid.com>
--
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