How to convert CV_8UC1 to normal int type?

Question

I'm having a problem with understanding the opencv types. I have matrix in CV_8UC1 type but how can I read value of elements in matrix?

I know that I have to use at method and then pas <here> my type but what type is CV_8UC1? 8 unsigned bit single channel doesn't tell me much.

Can i do something like this: unsigned int a = mat->at(0,0);

Answer 1

From the OpenCV Reference Manual:

CV_8UC1 – unsigned 8-bit single-channel data; can be used for grayscale image or binary image – mask.

So it's just single channel 8 bit grey scale with values 0..255.

Answer 2

Here are few examples:

• P.s stackoverflow sometimes doesnt allow to use >< characters so I will use } and { instead

.

// Char single channel matrix
Mat M1 = mat::ones(3,5,CV_8UC1);
int Val = M1{unsigned char}.at(2,3);

// int 16 matrix with 3 channels (like RGB pixel)
Mat M2 = mat::ones(3,5,CV_16UC(3));
__int16* Pix = &M2.at<__int16>(2,3);
Val = Pix[0] + Pix[1] + Pix[2];

// Example of how to find the size of each element in matrix
switch ( (M.dataend-M.datastart) / (M.cols*M.rows*M.channels())){
case sizeof(char):
     printf("This matrix has 8 bit depth\n");
     break;
case sizeof(double):
     printf("This matrix has 64 bit depth\n");
     break;
}

//Example of how to build dynamically a Matrix with desired depth
int inline CV_BUILD_MATRIX_TYPE(int elementSize, int nChannels){
    // Determine type of the matrix 
    switch (elementSize){
    case sizeof(char):
         return CV_8UC(nChannels);
         break;
    case (2*sizeof(char)):
         return CV_16UC(nChannels);
         break;
    case sizeof(float):
         return CV_32FC(nChannels);
         break;
    case sizeof(double):
         return CV_64FC(nChannels);
         break;
    }
    return -1;
}

Mat M = Mat::zeros(2,3,CV_BUILD_MATRIX_TYPE(4,2)); 
// builds matrix with 2 channels where each channel has depth of 4 bytes (float or __int32). As you wish