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

[Michael Niedermayer]

Hi

On Thu, Jul 19, 2007 at 10:52:38AM +0200, Benoit Fouet wrote:
> 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 ?

fine with me

[...]
-- 
Michael     GnuPG fingerprint: 9FF2128B147EF6730BADF133611EC787040B0FAB

Democracy is the form of government in which you can choose your dictator
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