OpenCV.2.Computer.Vision.Application.Programming.Cookbook--Scanning an image with pointers

#include<opencv2\opencv.hpp>

void colorReduce(cv::Mat &image, int div=64) 
{  
	int nr= image.rows; // number of rows  
	int nc= image.cols * image.channels(); // total number of elements per line  
	for (int j=0; j<nr; j++)
	{  
		// get the address of row j
		//ptr:It is a template method that returns the address of row number j:
		uchar* data= image.ptr<uchar>(j);  
		for (int i=0; i<nc; i++) 
		{  
			//we could have equivalently used pointer arithmetic to move from column to column

			// process each pixel ---------------------

			//data[i]= data[i]/div*div + div/2;  

			//data[i]= data[i]-data[i]%div + div/2;

			// mask used to round the pixel value
			int n=6;
			uchar mask= 0xFF<<n;
			data[i]=(data[i]&mask) + div/2;
			// e.g. for div=16, mask= 0xF0

			// end of pixel processing ----------------
		}                    
	}  
}  

int main(int argc,char* argv[])
{
	cv::Mat pImg;

	pImg=cv::imread("lena.jpg");
	cv::namedWindow("Image");
	cv::imshow("Image",pImg);

	colorReduce(pImg);

	
	cv::namedWindow("pImg");
	cv::imshow("pImg",pImg);

	cv::waitKey(0);

	cv::destroyWindow("Image");
	return 0;
}

color reduction is achieved by taking advantage of an integer division that floors the division result to the nearest lower integer:       

data[i]= data[i]/div*div + div/2;

The reduced color could have also been computed using the modulo operator which brings us to the nearest multiple of div (the 1D reduction factor):    

data[i]= data[i] – data[i]%div + div/2;
But this computation is a bit slower because it requires reading each pixel value twice.

Another option would be to use bitwise operators. Indeed, if we restrict the reduction factor to a power of 2, that is, div=pow(2,n), then masking the first n bits of the pixel value would give us the nearest lower multiple of div. This mask would be computed by a simple bit shift:
// mask used to round the pixel value
uchar mask= 0xFF<<n;

 // e.g. for div=16, mask= 0xF0

The color reduction would be given by:    

data[i]= (data[i]&mask) + div/2;
In general, bitwise operations lead to very efficient code, so they could constitute a powerful alternative when efficiency is a requirement.