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 valueint n= static_cast<int>(log(static_cast<double>(div))/log(2.0));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.

In our color reduction example, the transformation is directly applied to the input image,which is called an in-place transformation.However, in some applications, the user wants to keep the original image intact. The user would then be forced to create a copy of the image before calling the function. Note that the easiest way to create an identical deep copy of an image is to call the clone method

cv::Mat pImgClone=pImg.clone();colorReduce0(pImgClone);

This extra overload can be avoided by defining a function that gives the option to the user to either use or not use the in-place processing.

void colorReduce12(const cv::Mat &image, // input image                    cv::Mat &result,      // output image                   int div=64) {  int nr= image.rows; // number of rows  int nc= image.cols ; // number of columns  // allocate output image if necessary  result.create(image.rows,image.cols,image.type());  // created images have no padded pixels  nc= nc*nr;   nr= 1;  // it is now a 1D array  int n= static_cast<int>(log(static_cast<double>(div))/log(2.0));  // mask used to round the pixel value  uchar mask= 0xFF<<n; // e.g. for div=16, mask= 0xF0  for (int j=0; j<nr; j++) {  uchar* data= result.ptr<uchar>(j);  const uchar* idata= image.ptr<uchar>(j);  for (int i=0; i<nc; i++) {  *data++= (*idata++)&mask + div/2;  *data++= (*idata++)&mask + div/2;  *data++= (*idata++)&mask + div/2;  } // end of row                   }  }  

int main(int argc,char* argv[]){cv::Mat pImg,pImg1;pImg=cv::imread("lena.jpg");//cv::namedWindow("Image");cv::imshow("Image",pImg);//salt(pImg,300);//cv::Mat pImgClone=pImg.clone();//colorReduce0(pImgClone);colorReduce12(pImg,pImg1);//cv::namedWindow("pImg1");cv::imshow("pImg1",pImg1);cv::waitKey(0);//cv::destroyWindow("Image");//cv::destroyWindow("pImgClone");//cv::destroyAllWindows();return 0;}

实验结果：