Re: Edge Detection in C++

"DaveWilson" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Well that got it working :) thankyou for the help. but your right I
> would like to understand this a little better. I now understand the
> need for a second buffer. But am still a little in the dark about the
> actual operation and how it aplies to the code?
> Most of the code i undrstood but was a little unclear on the following
> sections:
>

Just taking these in a different order from your message. The following
loop -

>      // apply filter at this xy
>      nXSum = nYSum = 0;
>      for (i = -1; i <= 1; i++)
>      {
>       for (j = -1; j <= 1; j++)
>       {
>        ........

is looking at the 3x3 pixels around the current (x,y) - so the variable "i"
scans over the x dimension from one column left to one column right, and the
variable "j" scans over the y dimension from one row up to one row down.
Within these loops, nNeighX and nNeighY contain the coordinates of each of
the 3x3 pixels in turn -

>      nValue = my_image.raw_image[(nNeighY * my_image.width) +
> nNeighX];

Then the sobel operator is applied by taking the matching coefficient from
the static arrays nSobelX and nSobelY, which contain the h1 and h3 masks.
These are both 3x3 arrays and are accessed by zero-based index (like C
always does), so I add 1 to "i" and "j" (which range from -1 to +1) to get
the correct entry (in the range 0 to 2) and then accumulate the result for
both x and y -

>      nXSum += nValue * nSobelX[j + 1][i + 1];
>      nYSum += nValue * nSobelY[j + 1][i + 1];
>

After applying the masks over the whole 3x3 area, nXSum and nYSum will hold
the (approximated) gradient in the x and y directions respectively. So the
magnitude of this gradient vector is used to derive the edge strength -

>      dResult = (double)((nXSum * nXSum) + (nYSum * nYSum));
>      if (dResult > 0.0) dResult = sqrt(dResult);

The edge strength is scaled and stored in my work buffer before proceeding
to the next (x,y) -

>      pWork[(y * my_image.width) + x] = dResult / dScale;

Does this make it any clearer?

Regards,
Roger

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EMail: roger dot rowland at rmrsystems dot co dot uk