Opencv in memory Mat representation

Question

I know that in memory opencv represents Mat objects as one big array. So if I have 3 channels mat of dimension 200x200 then In memory it will store this mat in an array of size

Answer 1

As you can see in the data layout reported on the documentation, you can access the values like:

for(int r=0; r<rows; ++r){
    for(int c=0; c<cols; ++c){
        for(int ch=0; ch<nchannels; ++ch){
            double val = array[(nchannels*mat.step*r) + (nchannels*c) + ch];
        }
    }
}

Answer 2

You can get values in array like this:

double val = array+ (r*mat.step)+c*mat.channels()+ch;

Answer 3

I found other answers a bit confusing: mat.step is the size of a row in bytes, not in (double) elements, and it does already take into consideration the number of channels. To access val you should use:

double* array = (double*) mat.data; // was (double) mat.data in the question
double value = array[ ((mat.step)/mat.elemSize1())*c+mat.channels()*r+ch]; // (mat.step)/mat.elemSize1() is the actual row length in (double) elements

You can verify this and other approaches comparing them with the .at<> operator as follows:

#include <iostream>
#include <opencv2/core.hpp>

using namespace cv;
using namespace std;

int main()
{
const int w0=5;
const int h=3;
const int w=4;
double data[w0*h*3];
for (int y=0; y<h; y++)
    for (int x=0; x<w0; x++)
        for (int ch=0; ch<3; ch++)
            data[3*(x+w0*y)+ch]=1000+100*(y)+10*(x)+ch;

Mat m0(h,w0,CV_64FC3, data);
Rect roi(0,0,w,h);
Mat mat=m0(roi);

int c=3, r=2, ch=1;
Vec3d v = mat.at<Vec3d>(r,c);
cout<<"the 3 channels at row="<<r<<", col="<<c<<": "<<v<<endl;
double* array= (double*) mat.data;
double expect = 1000+100*r+10*c+ch;
double value= array[ ((mat.step)/mat.elemSize1())*r+mat.channels()*c+ch];
cout<<"row="<<r<<", col="<<c<<", ch="<<ch<<": expect="<<expect<<", value="<<value<<endl;
return  0;
}