Question

编辑：
指针/引用的用法可能是错误的，但是，如果任何cv :: Mat变量在均衡块外声明（在我的情况下，它们是其成员的类的定义），则也会发生OpenCV行为。

我在代码中使用的逻辑与所描述的here相同，但是结果却得到了奇怪的黑白图像，请参见original和result。我在Visual Studio 2017 15.8.8中将OpenCV 4.0.0与C ++一起使用。以前的声明：

cv::Mat *equalized_image;
cv::Mat &original_image = cv::Mat(cv::imread(file_path));

其中file_path是std :: string。均衡：

if (original_image.channels() >= 3) {
    // convert to YCrCb colourspace for luminance channel
    cv::cvtColor(original_image, *equalized_image, cv::COLOR_BGR2YCrCb);

    // split image by channels
    std::vector<cv::Mat> channels;
    cv::split(*equalized_image, channels);

    // equalize only the histogram of the luminance channel
    cv::equalizeHist(channels[0], channels[0]);

    // merge back
    cv::merge(channels, *equalized_image);

    // convert back to BGR colourspace
    cv::Mat result;
    cv::cvtColor(*equalized_image, result, cv::COLOR_YCrCb2BGR);

    cv::namedWindow("Equalized");
    cv::imshow("Equalized", result);
}

Answer 1

我很惊讶您的代码被编译。您使用的逻辑很好，但是您编码的指针（* equalized_image）和引用（＆original_image）令人困惑。如果您在发布模式下运行，它可以做一些有趣的事情来初始化内存并使工作正常。始终先在调试模式下检查代码。

将代码更改为此，它应该可以按预期工作：

cv::Mat equalized_image;
cv::Mat original_image = cv::imread(file_path);

// convert to YCrCb colourspace for luminance channel
cv::cvtColor(original_image, equalized_image, cv::COLOR_BGR2YCrCb);

// split image by channels
std::vector<cv::Mat> channels;
cv::split(equalized_image, channels);

// equalize only the histogram of the luminance channel
cv::equalizeHist(channels[0], channels[0]);

// merge back
cv::merge(channels, equalized_image);

// convert back to BGR colourspace
cv::Mat result;
cv::cvtColor(equalized_image, result, cv::COLOR_YCrCb2BGR);

cv::namedWindow("Equalized");
cv::imshow("Equalized", result);
waitKey(0);

OpenCV EqualizeHist（）从彩色图像创建黑白图像

1 个答案: