使用3通道使用opencv创建直方图图表

Question

我试图绘制3通道图像的直方图，以及“opencv_tut.exe中0x752e812f处未处理的异常：Microsoft C++异常：内存位置0x0019ef5c处的cv :: Exception”这个错误。甚至不能调试它继续进入组装。

问题是在直方图中的线条绘制部分，其他人只是正常工作。

下面是代码：

空隙colorHistogram（）{

Mat image = imread("c:/aaa.jpg"); 
vector<Mat> bgr_planes; 
split(image, bgr_planes); 

if(!image.data){ 

    cout<<"No image"<<endl; 

} 

int histSize = 256; 
float range[] = {0, 256}; 
const float* histRange = {range}; 

Mat b_hist, g_hist, r_hist; 

calcHist(&bgr_planes[0], 1, 0, Mat(), b_hist, 1, &histSize, &histRange, true, false); 
calcHist(&bgr_planes[1], 1, 0, Mat(), g_hist, 1, &histSize, &histRange, true, false); 
calcHist(&bgr_planes[2], 1, 0, Mat(), r_hist, 1, &histSize, &histRange, true, false); 

int hist_w = 512; 
int hist_h = 400; 

int bin_w = cvRound((double) hist_w/histSize); 

Mat histImage(hist_h, hist_w, CV_8UC3, Scalar(0,0,0)); 

normalize(b_hist, b_hist, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 
normalize(g_hist, g_hist, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 
normalize(r_hist, r_hist, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 

for(int i=0; i<histSize; i++){ 

    line(histImage, Point(bin_w*(i-1), hist_h - cvRound(b_hist.at<float>(i-1))) , Point(bin_w*(i), hist_h - cvRound(b_hist.at<float>(i))), Scalar(255, 0, 0), 2, 8, 0 ); 
    line(histImage, Point(bin_w*(i-1), hist_h - cvRound(g_hist.at<float>(i-1))), Point(bin_w*(i), hist_h - cvRound(g_hist.at<float>(i))), Scalar(0, 255, 0), 2, 8, 0 ); 
    line(histImage, Point(bin_w*(i-1), hist_h - cvRound(r_hist.at<float>(i-1))) , Point(bin_w*(i), hist_h - cvRound(r_hist.at<float>(i))), Scalar(0, 0, 255), 2, 8, 0 ); 

} 

namedWindow("Histogram", CV_WINDOW_AUTOSIZE); 
imshow("Histogram", histImage); 

}

Answer 1

大概任性指针或超越所分配的存储器的边界。不要以为你以前的代码是正确的;像这样的问题有时可能会导致更早出现，并且只会在稍后出现。

在任何情况下，看来你在这里修改的代码：http://docs.opencv.org/doc/tutorials/imgproc/histograms/histogram_calculation/histogram_calculation.html

的主要区别在于你的for循环从0到HISTSIZE和他们从1到HISTSIZE。由于该索引用于计算循环中的阵列位置，所以我猜这就是你的问题。