RAW图转换为bmp图

Question

RAW图转换为bmp图，看不太懂转换过程....

Answer 1

 
/***********************************************************
打开raw图像，并在opencv中显示。需要知道原始图像的分辨率 (8221*7441)；
和数据类型为 float，利用fopen和memcpy内存拷贝赋值到opencv中并显示！！！
@zhou 2020/1/8
************************************************************/
#include<iostream>
#include<opencv2/opencv.hpp>
using namespace std;
using namespace cv;
int main()
{
    int height = 7441;//原始图像的高
    int width = 8221;//原始图像的宽
    /*int height = 7441;
    int width = 8221;*/
    FILE *fp = NULL; //定义指针
    fp = fopen("1.raw", "rb+");
    float *data = (float *)malloc(width*height * sizeof(float)); //内存分配，new，malloc都行
    fread(data, sizeof(float), width*height, fp); //在缓存中读取数据
    cv::Mat img;
    int bufLen = width*height;  //定义长度
    img.create(height, width, CV_32FC1);//创建Mat
    memcpy(img.data, data, bufLen * sizeof(float)); //内存拷贝
    /*
    //遍历图像的每个像素点
    for (int row = 0; row < img.rows; row++)
        {
            for (int col = 0; col < img.cols; col++)
                {
                    int d = img.at<float>(row, col);
                    printf("%d,%d,的值为：%d\n", row, col, d);
                }
        }
*/
    //图像(3678,535)处的像素值
    int pixel = img.at<float>(3678, 635);
    printf("%d,%d,的像素值为：%d\n", 3678, 635, pixel);
    //normalize(img,img, 1.0,255.0, NORM_INF);
    double minv = 0.0, maxv = 0.0;
    double* minp = &minv;
    double* maxp = &maxv;
    minMaxIdx(img, minp, maxp);//找到最大和最小的像素值
    cout << "最大的像素值为：" << *minp << endl;
    cout << "最大的像素值为：" << *maxp << endl;
    //img.convertTo(img, CV_8UC1, 255, 0); //归一化为0,255
    namedWindow("img", 0);//窗口可以缩放
    imshow("img", img); //显示
    //imwrite("out.jpg", img); //保存为jpg
    waitKey(0);
    return 0;
 
}