【OpenCV】三种方式操作图像像素

OpenCV中，有3种访问每个像素的方法：使用at方法、使用迭代器方法、使用指针

运行如下程序后可以发现使用at方法速度最快。

代码如下：

//操作图像像素
#include <opencv2/opencv.hpp>
#include <iostream>
#include <cstdio>
using namespace std;
using namespace cv;

void colorReduce(Mat &a, Mat &b, int div) {//at方法
    b = a.clone();
    int row = b.rows, col = b.cols;
    for(int i = ; i < row; ++i) {
        for(int j = ; j < col; ++j) {
            b.at<Vec3b>(i, j)[] = b.at<Vec3b>(i, j)[] / div * div + div/;
            b.at<Vec3b>(i, j)[] = b.at<Vec3b>(i, j)[] / div * div + div/;
            b.at<Vec3b>(i, j)[] = b.at<Vec3b>(i, j)[] / div * div + div/;
        }
    }
}
void colorReduce1(Mat &a, Mat &b, int div) {//指针
    b = a.clone();
    int row = b.rows, col = b.cols * b.channels();
    uchar *c = b.data;
    for(int i = ; i < row; ++i) {
        for(int j = ; j < col; ++j) {
            *c = *c / div * div + div / ;
            c++;
        }
    }
}
void colorReduce2(Mat &a, Mat &b, int div) {//迭代器
    b = a.clone();
    Mat_<Vec3b>::iterator it = a.begin<Vec3b>();
    Mat_<Vec3b>::iterator itend = a.end<Vec3b>();
    Mat_<Vec3b>cimg = b;
    Mat_<Vec3b>::iterator itout = cimg.begin();
    Mat_<Vec3b>::iterator itoutend = cimg.end();
    for(; it != itend; it ++,itout++) {
        (*itout)[] = (*it)[]/div * div + div / ;
        (*itout)[] = (*it)[]/div * div + div / ;
        (*itout)[] = (*it)[]/div * div + div / ;
    }
}
int main() {
    Mat img, res, temp;
    img = imread("PIC1.jpg");
    if(!img.data) {cout<<"read image file wrong!"<<endl; getchar(); return ;}
    //cout << "height = " << img.size().height << ",width = " << img.size().width << endl;
    getchar();
    imshow("PIC", img);

    double t1=(double)getTickCount();  

    //colorReduce(img, res, 128);//at
    colorReduce1(img, res, );//指针
    //colorReduce2(img, res, 128);//迭代器

    double t2=(double)getTickCount();
    double time = (t2-t1)/getTickFrequency()*;
    cout<<"Time = " << time << "ms" <<endl;  

    namedWindow("Picture");
    imshow("Picture", res);

    waitKey();
    return ;
}