[x264-devel] Doubts in computing integral image

[Aishwarya]

Hi

Thanks for clarifying the doubts BugMaster.


On Fri, Oct 14, 2016 at 11:21 AM, BugMaster <BugMaster at narod.ru> wrote:

> On Fri, 14 Oct 2016 10:11:16 +0530, Aishwarya wrote:
> > ...
>
>  >> static void integral_init8v( uint16_t *sum8, intptr_t stride )
>  >> {
>  >>     for( int x = 0; x < stride-8; x++ )
>  >>         sum8[x] = sum8[x+8*stride] - sum8[x];
>  >> }
>  >> In the above function why the loop is only upto stride - 8? As far
>  >> as I understood, since it is for vertical, loop could go well till
> stride.
> >
> >  Here stride - 8 is correct same as in integral_init4v for sum8 because
> >  in integral_init8h we calculate sum only up to stride - 8:
> >
> >  static void integral_init8h( uint16_t *sum, pixel *pix, intptr_t stride
> )
> >  {
> >      int v = pix[0]+pix[1]+pix[2]+pix[3]+pix[4]+pix[5]+pix[6]+pix[7];
> >      for( int x = 0; x < stride-8; x++ )
> >      {
> >          sum[x] = v + sum[x-stride];
> >          v += pix[x+8] - pix[x];
> >      }
> >  }
> >
>
> >
> > I agree that in init8h we need only till stride-8, as we keep
> > adding pix[x+8]. But that is not the case with init8v.
> > In init8v we are adding one pixel with a pixel which lies below a
> > few rows. So  I think we can go till stride.
> > Please clarify.
> >
>
> In init8v we use as source not the pix[] but sum[] which was result
> of init8h and which we filled only up to stride-8.
>
> > Here what I mean is that while computing sum4, the loop should go
> > till stride, and while computing sum8, it should go till stride - 4
> > (since in sum8, sum8[x+4] is accessed).
> >
>
> Same as above in init4v we use as source not pix[] but sum[] which was
> result of init4h and which we filled only up to stride-4. So it looks
> logical to use stide-4 for sum4 and stride-8 for sum8 (since in sum8,
> sum8[x+4] is accessed).
>
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