#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.