Issue

This Content is from Stack Overflow. Question asked by Maximiliano Gaggini

In the problem description, we were told to take outer pixels as having a value of “0”.
I discriminate de outer pixels so as to do nothing with them, they won´t add a value, because multiplying by 0 gets 0.
I can´t seem to find where i went wrong. Can somebody give me a clue. Thanks a lot.
here is my code

```````// Detect edges
void edges(int height, int width, RGBTRIPLE image[height][width])
{
// Make copy
RGBTRIPLE copy[height][width];
for (int i = 0; i < height; i++)
{
for (int j = 0; j < width; j++)
{
copy[i][j] = image[i][j];
}
}
double gxB, gxR, gxG, gyR, gyB, gyG;
int GX[3][3] = {{-1, 0, 1}, {-2, 0, 2}, {-1, 0, 1}};
int GY[3][3] = {{-1, -2, -1}, {0, 0, 0}, {1, 2, 1}};
// Loop through pixels
for (int i = 0; i < height; i++)
{
for (int j = 0; j < width; j++)
{
gxB = 0;
gxR = 0;
gxG = 0;
gyR = 0;
gyB = 0;
gyG = 0;
// Loop through 3x3 box
for (int row = (i - 1); row < (i + 2); row++)
{
for (int col = (j - 1); col < (j +2); col++)
{
// Discriminate border pixels
if (exists(row, col, height, width))
{
gxR += copy[row][col].rgbtRed * GX[row][col];
gxG += copy[row][col].rgbtGreen * GX[row][col];
gxB += copy[row][col].rgbtBlue * GX[row][col];

gyR += copy[row][col].rgbtRed * GY[row][col];
gyG += copy[row][col].rgbtGreen * GY[row][col];
gyB += copy[row][col].rgbtBlue * GY[row][col];
}
else
{
gxR += 0.0;
gxG += 0.0;
gxB += 0.0;

gyR += 0.0;
gyG += 0.0;
gyB += 0.0;
}
}
}
image[i][j].rgbtRed = trimm(round(sqrt((gxR * gxR) + (gyR * gyR))));
image[i][j].rgbtGreen = trimm(round(sqrt((gxG * gxG) + (gyG * gyG))));
image[i][j].rgbtBlue = trimm(round(sqrt((gxB * gxB) + (gyB * gyB))));
}
}

return;
}

bool exists(int ROW, int COL, int H, int W)
{
if (((ROW >= 0) && (ROW < H)) && ((COL >= 0) && (COL < W)))
{
return true;
}
else
{
return false;
}

}

int trimm(int x)
{
if (x < 256)
{
return x;
}
else
{
return 255;
}
}`
``````

Solution

This question is not yet answered, be the first one who answer using the comment. Later the confirmed answer will be published as the solution.

```