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= 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：