C++ – Show Vec3b pixel values in OpenCV Mat image

Assuming the type is CV_8UC3 you would do this:

for(int i = 0; i < foo.rows; i++)
{
    for(int j = 0; j < foo.cols; j++)
    {
        Vec3b bgrPixel = foo.at<Vec3b>(i, j);

        // do something with BGR values...
    }
}

Here is the documentation for Vec3b. Hope that helps! Also, don't forget OpenCV stores things internally as BGR not RGB.

EDIT :
For performance reasons, you may want to use direct access to the data buffer in order to process the pixel values:

Here is how you might go about this:

uint8_t* pixelPtr = (uint8_t*)foo.data;
int cn = foo.channels();
Scalar_<uint8_t> bgrPixel;

for(int i = 0; i < foo.rows; i++)
{
    for(int j = 0; j < foo.cols; j++)
    {
        bgrPixel.val[0] = pixelPtr[i*foo.cols*cn + j*cn + 0]; // B
        bgrPixel.val[1] = pixelPtr[i*foo.cols*cn + j*cn + 1]; // G
        bgrPixel.val[2] = pixelPtr[i*foo.cols*cn + j*cn + 2]; // R

        // do something with BGR values...
    }
}

Or alternatively:

int cn = foo.channels();
Scalar_<uint8_t> bgrPixel;

for(int i = 0; i < foo.rows; i++)
{
    uint8_t* rowPtr = foo.row(i);
    for(int j = 0; j < foo.cols; j++)
    {
        bgrPixel.val[0] = rowPtr[j*cn + 0]; // B
        bgrPixel.val[1] = rowPtr[j*cn + 1]; // G
        bgrPixel.val[2] = rowPtr[j*cn + 2]; // R

        // do something with BGR values...
    }
}

Here you can see some of the possibilities for fast element access.

But if you want to do it your way, you need to add a bracket. Otherwise you index computation is not correct:

for(int i=0; i<A.rows; i++){
   for(int j=0; j<A.cols; j++){
           A.data[A.channels()*(A.cols*i + j) + 0] = 0;
           A.data[A.channels()*(A.cols*i + j) + 1] = 0;
           A.data[A.channels()*(A.cols*i + j) + 2] = 0;
    }
} 

But the layout of the memory is not guaranteed to be contiguous due to padding. So according to this you should rather use a formula like this:

for(int i=0; i<A.rows; i++){
   for(int j=0; j<A.cols; j++){
           A.data[A.step[0]*i + A.step[1]* j + 0] = 0;
           A.data[A.step[0]*i + A.step[1]* j + 1] = 0;
           A.data[A.step[0]*i + A.step[1]* j + 2] = 0;
    }
}