[FFmpeg-cvslog] r9733 - trunk/libavformat/os_support.c

[Benoit Fouet]

Uoti Urpala wrote:
> On Thu, 2007-07-19 at 10:17 +0200, Benoit Fouet wrote:
>   
>> M?ns Rullg?rd wrote:
>>     
>>>>      add1 = atoi(pch);
>>>>      pch = strpbrk(pch,".");
>>>> -    if (pch == 0 || ++pch == 0) return 0;
>>>> +    if (!pch) return 0;
>>>>      add2 = atoi(pch);
>>>>         
>
>   
>>> This is wrong.  The ++ is needed.  Whoever wrote this code was very
>>> confused.
>>>   
>>>       
>> then the right thing to do would be:
>> if(!pch) return 0;
>> addX = atoi(++pch);
>>
>> right ? tome, the check of ++pch is really unneeded
>>     
>
> The ++ is needed to move over the '.', but yes check against 0 will
> always be false. The next character of the string (there must be one if
> the string is 0-terminated and the current character is not 0) will not
> be at address 0.
>
> I guess it was meant to be a check against the pointed-to character
> being the terminating 0. However even with that the error checking in
> that function would be really weak - it'll accept addresses such as
> "1.12foo.bar.something.32000.1.1.2" (I think that would be parsed the
> same as "1.12.0.0").
>
>   

so what about using something like
if( sscanf(pch, "%d.%d.%d.%d", &add1, &add2, &add3, &add4) != 4 )
  return 0;

instead of the strpbrk stuff ?

-- 
Ben
Purple Labs S.A.
www.purplelabs.com