>>>>> "metiu" == metiu uitem <metiu.uitem@gmail.com> writes:

    metiu> Hi,
    metiu> I'm building a quantization table to use with the IMG_quantize function
    metiu> on a TI DSP to quantize the DCT of an image.
    metiu> The function basically does this:

    metiu> short quantValue (short v)
    metiu> {
    metiu>     int q, round;

    metiu>     round = 1 << (shift -  1)
    metiu>     q = (v * recip + round) >> shift

    metiu>     return (short) q;
    metiu> }

    metiu> I used the algorithm to find the reciprocal on Jones page:
    metiu> http://www.cs.uiowa.edu/~jones/bcd/divide.html

    metiu> The first problem I found is that using the "optimal" reciprocal leads
    metiu> to a "negative" short, which would give -v/d instead of v/d. I arranged
    metiu> to have a reciprocal with a leading 0 to avoid that.

What is the reciprocal you are trying to find?

Ray

Hi,
I'm building a quantization table to use with the IMG_quantize function
on a TI DSP to quantize the DCT of an image.
The function basically does this:

short quantValue (short v)
{
    int q, round;

    round = 1 << (shift -  1)
    q = (v * recip + round) >> shift

    return (short) q;
}

I used the algorithm to find the reciprocal on Jones page:
http://www.cs.uiowa.edu/~jones/bcd/divide.html

The first problem I found is that using the "optimal" reciprocal leads
to a "negative" short, which would give -v/d instead of v/d. I arranged
to have a reciprocal with a leading 0 to avoid that.

My problem is: I need the rounded down division, not the rounded one.
To do so, I'm doing:
if (v > d/2) v -= d/2;
else if (v < -d/2) v += d/2;

but that's very unefficient.

I think I could find a reciprocal that automatically rounds down, but I
can't think of a (simple) way to do that...
Any help/pointer?

Thanks,
M