[FFmpeg-devel] [PATCH] Implement in lavc a flag which makes avcodec_open() to choose the best framerate

[Stefano Sabatini]

On date Sunday 2008-09-21 22:16:56 +0200, Michael Niedermayer encoded:
> On Sun, Sep 21, 2008 at 06:52:23PM +0200, Stefano Sabatini wrote:
> > On date Sunday 2008-09-21 14:09:56 +0200, Michael Niedermayer encoded:
> > > On Sun, Sep 21, 2008 at 12:52:57PM +0200, Stefano Sabatini wrote:
> > [...]
> > > > Index: libavutil/rational.c
> > > > ===================================================================
> > > > --- libavutil/rational.c	(revision 15372)
> > > > +++ libavutil/rational.c	(working copy)
> > > > @@ -101,3 +101,81 @@
> > > >  
> > > >      return a;
> > > >  }
> > > > +
> > > > +int av_is_nearest_q(AVRational q, AVRational q1, AVRational q2)
> > > > +{
> > > > +    if (av_cmp_q(q1, q) * av_cmp_q(q, q2) > 0) {
> > > > +        /* q1 and q2 on the opposite sides of q */
> > > > +        /* N/D is q, A/B is the median between q1 and q2, A/B > N/D <=> A*D/B > N */
> > > 
> > > > +        int64_t x = av_rescale(q1.num * (int64_t)q2.den + q2.num * (int64_t)q1.den, q.den, 2 * (int64_t)q1.den * q2.den);
> > > 
> > > the rounding is not correct
> > 
> > I see that the type of rounding choosen may change the result of the
> > comparation if the result isn't exact, but I can't currently figure
> > out which is the correct one to apply.
> > 
> > Can you suggest which should I use?
> 
> to test A*D/B > N
> the left side should be rounded toward inf

Sorry to insist on this, what if we have:

X = A*D/B
X'= rnd(A*D/B, up), X' > X
X < N but X' > N

Is this possible at all?

> > > > +        if (((x > q.num ? 1 : x < q.num ? -1 : 0) * av_cmp_q(q2, q1)) > 0)
> > > > +            return 1;
> > > > +    }
> > > 
> > > > +    /* q1 and q2 on the same side of q */
> > > > +    else if (av_cmp_q(q, q1) * av_cmp_q(q1, q2) > 0)
> > > 
> > > i dont see why these special cases would be needed
> > 
> > The first condition:
> > cmp(M, q) * cmp(q2, q1) == 1 
> > 
> > catches both the conditions:
> > q1 ... q ... M ... q2
> > M > q && q2 > q1
> > and 
> > q2 ... M ... q ... q1
> > M < q && q2 < q1
> > 
> > (x > q.num ? 1 : x < q.num ? -1 : 0) corresponds to cmp(M, q).
> > 
> > The second condition:
> > cmp(q, q1) * cmp(q1, q2) ==  1
> > cathes both the conditions:
> > q2 ... q1 ... q
> > q > q1 && q1 > q2
> > and 
> > q ... q1 ... q2
> > q < q1 && q1 < q2
> > 
> > I can't see how to simplify this...
> 
> what is the problem with the code for the first case handling the second?

Uh?

1 if (q1 and q2 on opposite sides of q)
2    if (q1 nearest than q2)  // opposite sides context
3       return 1;
4 else // q1 and q2 are on the same side of q
5    if (q1 nearest than q2) // same side context)
6       return 1;
7 return 0;

Checks at points 1, 2 and 5 test different things, so I don't see how
it would be possible to reuse the code of one test for another
one. You we could make them parametric, but I'm not sure that would
simplify the code, rather it would make it more obfuscated.

But I feel like I'm misunderstanding your point...

Regards.
-- 
FFmpeg = Fantastic and Fantastic Merciful Portentous Ecumenical Governor
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