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Date: Tue, 20 Nov 2007 13:39:05 +0800
From: WANG Cong <xiyou.wangcong@gmail.com>
To: "H. Peter Anvin" <hpa@zytor.com>
Cc: WANG Cong <xiyou.wangcong@gmail.com>, zhengyi <goodmenkernel@gmail.com>,
       linux-kernel@vger.kernel.org
Subject: Re: Is there any word about this bug in gcc ?
Message-ID: <20071120053904.GG2472@hacking>
Reply-To: WANG Cong <xiyou.wangcong@gmail.com>
References: <20071120.101342.85695496.goodmenzy@gmail.com> <20071120041600.GC2472@hacking> <47426C54.9020009@zytor.com>
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On Mon, Nov 19, 2007 at 09:10:44PM -0800, H. Peter Anvin wrote:
>WANG Cong wrote:
>>On Tue, Nov 20, 2007 at 10:13:42AM +0800, zhengyi wrote:
>>>Is there any relevance to the kernel ?
>>>
>>>I found the folowing code here:
>>>http://linux.solidot.org/article.pl?sid=07/11/19/0512218&from=rss
>>>
>>>-------------------------------------------------------------------
>>>int main( void )
>>>{
>>> int i=2;
>>> if( -10*abs (i-1) == 10*abs(i-1) )
>>>   printf ("OMG,-10==10 in linux!\n");
>>> else
>>>   printf ("nothing special here\n") ;
>>>
>>> return 0 ;
>>>}
>>
>>I think no. It is considered a bug in abs(), kernel, of course,
>>doesn't use glibc's abs().
>>
>
>Wrong.
>
>abs() is internal to gcc, and the above is optimized out at compile 
>time, so any user of abs() as a function at all is vulnerable.

This is an urgent bug, I think.

And you mean abs() is not in glibc, then where is it? Built in gcc?
And what's more, why not put it in glibc?

Thanks.

>
>However, the Linux kernel defines abs() as a macro:
>
>#define abs(x) ({                               \
>                int __x = (x);                  \
>                (__x < 0) ? -__x : __x;         \
>        })
>
>... which means gcc never sees it.  So the kernel isn't affected, 
>because it doesn't use *gcc's* abs().

Thanks for clarifying this!

Regards.


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