使用 C++ 将 RGBA 转换为 YUV422 格式 - 没有得到正确的输出

Question

我想将图像从 RGBA 格式转换为 YUV422 格式。下面是我的代码：

#include "opencv2/highgui/highgui.hpp"
#include "opencv2/imgproc/imgproc.hpp"
#include <iostream>
#include <stdio.h>
#include <stdlib.h>

using namespace cv;
using namespace std;

#define CLIP(X) ( (X) > 255 ? 255 : (X) < 0 ? 0 : X)
#define RGB2Y(R, G, B) CLIP(( (  66 * (R) + 129 * (G) +  25 * (B) + 128) >> 8) +  16)
#define RGB2U(R, G, B) CLIP(( ( -38 * (R) -  74 * (G) + 112 * (B) + 128) >> 8) + 128)
#define RGB2V(R, G, B) CLIP(( ( 112 * (R) -  94 * (G) -  18 * (B) + 128) >> 8) + 128)

int main(int argc, char *argv[])
{

    Mat in_img, in_RGB, out_yuv;
    in_img = imread(argv[1],1);
    if(!in_img.data)
    {
        printf("Failed to load the image ... %s\n!", argv[1]);
        return -1;
    }
    short imgwidth  = in_img.cols;
    short imgheight = in_img.rows;

    cvtColor(in_img, in_RGB, CV_BGR2RGB);  
    out_yuv.create(imgheight, imgwidth, CV_16U);

    imwrite("RGB.jpg",in_RGB);

    for(int i = 0; i< imgheight; i++)
        for(int j = 0; j< imgwidth; j = j+2) {
            unsigned int val = in_RGB.at<unsigned int>(i,j);

/*********** for 1st pixel ***********/
            unsigned int tmp = val;
            unsigned char B = (unsigned char)(((tmp<<16) >> 24) | 0x000000ff ); //Extracting B channel data

            tmp = val;
            unsigned char G = (unsigned char)(((tmp<<8) >> 24) | 0x000000ff ); //Extracting G channel data

            tmp = val;
            unsigned char R = (unsigned char)((tmp >> 24) | 0x000000ff );  //Extracting R channel data

            unsigned char Y1 = RGB2Y(R, G, B);
            unsigned char U = RGB2U(R, G, B);
            unsigned char V = RGB2V(R, G, B);

/*********** for 2nd pixel ***********/
            unsigned int val1 = in_RGB.at<unsigned int>(i,j+1);
            unsigned int tmp1 = val1;
            unsigned char B1 = (unsigned char)(((tmp1<<16) >> 24) | 0x000000ff );

            tmp1 = val1;
            unsigned char G1 = (unsigned char)(((tmp1<<8) >> 24) | 0x000000ff );

            tmp1 = val1;
            unsigned char R1 = (unsigned char)((tmp1 >> 24) | 0x000000ff );

            unsigned char Y2 = RGB2Y(R, G, B);

/********** writing into out image in U-Y1-V-Y2 format *********************/

        unsigned short p1 = ((U << 8) | Y1) ;
        unsigned short p2 = ((V << 8) | Y2) ;
        out_yuv.at<unsigned short>(i,j) = p1;
        out_yuv.at<unsigned short>(i,j+1) = p2;

    }
        imwrite("YUV422.png",out_yuv);
return 0;   
}

但是我没有得到正确的输出图像。我也尝试用 Y1-U-Y2-V 格式编写它，输出仍然相同。上面的代码可能有什么问题，或者是创建问题的输出图像格式（.png/.jpg）？（cv::imwrite 不允许以 .yuv 格式保存）

Answer 1

out_yuv.create(imgheight, imgwidth, CV_16U);
...
imwrite("YUV422.png",out_yuv);

这里有几个问题：

imwrite 考虑 CV_16U to be mono channel。
你不能在png里面写yuv。您必须转储原始数据，并使用特殊的 yuv 查看器，例如 raw yuvplayer 以确保您的数据正确。

编辑：
请注意，原始像素首字母缩略词有点大端，即最左边的值是内存中的最低有效字节（大多数时候，视频人们喜欢混乱）

RGBA 表示 R = RGB[0]、G = RGB[1]、B = RGB[2]、A = RGB[3]
YUYV 表示在缓冲区中 p 表示 p[0] 是 Y0，p[1] 是 U，p[2] 是 Y1，p[3] 是 V

你的轮班正好相反:)

编辑 2：如果你的图像仍然是乱码，你似乎愿意按照平面格式（YUV422P akaYV16）来写。平面格式意味着每个通道都在一个单独的、连续的平面上。对目标数据使用普通的旧缓冲区（unsigned char*，或std::vector<unsigned char>）更清晰。

unsigned char* out_yuv_ptr = new unsigned char[imgwidth*imgheight*2];
unsigned char* y_ptr = out_yuv_ptr;
unsigned char* u_ptr = y_ptr + imgwidth*imgheight;
unsigned char* v_ptr = u_ptr + imgwidth*imgheight/2;
...
*y_ptr = Y1;
++y_ptr
*y_ptr = Y2;
++y_ptr
*u_ptr = U;
++u_ptr;
*v_ptr = V;
++v_ptr;

使用 C++ 将 RGBA 转换为 YUV422 格式 - 没有得到正确的输出

Converting RGBA to YUV422 format using c++ - not getting proper output

c++

opencv

image

image-processing

color-channel