将32位rgb图像转换为8位rgb C++

Question

我想把w的uint32_t图像数据转换成uint8_t图像。我该如何转换图像。我想把轻子子的PIX转换成opencv Mat。

我想知道使用按位运算符的困难方法。像素被打包为AARRGGBB。

另外，我也想皈依

简历:Mat 8UC1至PIX。即8位单通道图像到32位图像。或者你能想出其他办法来解决这个问题。

Answer 1

以下是将cv::Mat转换为32位PIX格式的答案：

PIX* toPIX(const cv::Mat& img)
{
    if(img.empty())
        throw std::invalid_argument("Image is empty");

    //PIX has got 4 channels hence we need to add one
    //more channel to this image
    cv::Mat alphaImage;
    if(img.type() == CV_8UC3) {
        cv::cvtColor(img,alphaImage,cv::COLOR_BGR2RGBA);
    } else if(img.type() == CV_8UC1) {
        cv::Mat gray = img.clone();
        std::vector<cv::Mat> channelsPlusAlpha;

        //construct 4 channel by reapeating gray across 3 channels and filling alpha with constant value
        channelsPlusAlpha.push_back(gray);//channels[2]);//R
        channelsPlusAlpha.push_back(gray);//channels[1]);//G
        channelsPlusAlpha.push_back(gray);//channels[0]);//B
        channelsPlusAlpha.push_back(cv::Mat(img.size(),CV_8UC1,cv::Scalar::all(255)));//A
        cv::merge(channelsPlusAlpha,alphaImage);
    } else {
        throw std::logic_error("Image type undefined");
    }


    //Prepare PIX
    int w = alphaImage.cols;
    int h = alphaImage.rows;
    int bpl = alphaImage.step1();//bytes per line
    int bpp = alphaImage.channels();//bytes per pixel
    int top = 0; //x (0,0) or if we want to deal with sub image then those coordinates have to be fed
    int left = 0;//y
    uchar* image_data = alphaImage.data;

    //Create PIX
    PIX *pix = pixCreate(w,h,32);
    uint32_t* data = pixGetData(pix);
    int wpl = pixGetWpl(pix);//get the words per line

    const uint8_t* imagedata = image_data + top * bpl + left * bpp;

    for(int y=0; y < h; ++y) {
        const uint8_t* linedata = imagedata; // linescan
        uint32_t* line = data + y *wpl;
        for(int x =0; x < w; ++x) {
            line[x] =   (linedata[0] << 24) |
                        (linedata[1] << 16) |
                        (linedata[2] << 8)  |
                        linedata[3];

            linedata += 4;
        }
        imagedata += bpl;
    }

    return pix;
}

Answer 2

如果您想用数据获取一个4通道的RGBA映像，可以尝试将每个uint32_t转换为4个无符号字符号。有关如何进行此操作的讨论，请参见Converting an int into a 4 byte char array (C)。完成此操作后，只需使用所获得的数据创建一个类型为cv::Mat的新CV_8UC4。

Answer 3

uint32_t * imData = yourImageDataPointer;
uchar * reinterpretedData = (uchar*)imData;

cv::Mat cvImage = cv::Mat(h,w, CV_8UC4);

cv::Mat bgr; // displayable image without alpha channel

cv::cvtColor(cvImage, bgr, CV_RGBA2BGR);

cv::imshow("img", bgr);
cv::waitKey (0);

如果结果不是你所期望的，请尝试这个，并发布结果图像。