我正试图用opencv和c++写一个程序。我有一个图像,我试图得到(x, y)点的确定像素的饱和度值。我用下面的句子来做:
saturation_level = hsv_chanels[1].at<uchar>(x, y);
问题是程序构建正常,但当我尝试运行时,它有时工作正常,有时以以下错误终止:
分段故障:11
有人知道为什么会出现这个错误吗?我读到这个错误出现是因为我的计算机内存,但我不知道为什么它只出现有时。
编辑:我调用这个函数来查找单应字:
Mat ObtenHomografiaSuelo (vector <KeyPoint> keypoints1, vector <KeyPoint> keypoints2, Mat imagen1, Mat imagen2){
//*****************************************************************************
//Find homography mat
//*****************************************************************************
vector < Point2f > image_points[2];
int cont = 0;
vector<Mat> chanels_hsv1; //[0]->H, [1]->S, [2]->V
split( image1, chanels_hsv1 );
vector<Mat> chanels_hsv2;
split( image2, chanels_hsv2 );
for(vector<KeyPoint>::const_iterator it = keypoints1.begin(); it!= keypoints1.end(); ++it){
// Get the position of left keypoints
float x = (it->pt.x);
float y = (it->pt.y);
cout << "1" << endl;
float saturation_level = chanels_hsv1[1].at<uchar>(x, y);
cout << "2" << endl;
double max_level = 70.0;
cout << "3" << endl;
if ((y < camSize.height/4) && (saturatio_level < max_level) ){
cout << "1:" << endl;
waitKey (100);
cout << "y: " << y;
cout << " Saturation_Level: " << nivel_saturacion << endl;
image_points[0].push_back(Point2f(x,y));
cout << "done" << endl;
cont ++;
}
}
cont = 0;
for (vector<KeyPoint>::const_iterator it = keypoints2.begin(); it!=keypoints2.end(); ++it) {
// Get the position of left keypoints
float x = (it->pt.x);
float y = (it->pt.y);
float saturation_level = chanels_hsv2[1].at<uchar>(x, y);
double max_level = 70.0;
if ((y < (camSize.height)/4) && (saturation_level < max_level)){
cout << "2" << endl;
waitKey (100);
cout << "y: " << y;
cout << " Saturation_Level: " << nivel_saturacion << endl;
image_points[1].push_back(Point2f(x,y));
cont ++;
}
}
cout << "We are obtain: " << cont << " points to do the homography" << endl;
waitKey();
Mat H;
H = Mat::zeros(4, 4, CV_64F);
if (cont < 4) {
cout << "Few points to do the homography" << endl;
}
else{
if (image_points[0].size() > image_points[1].size()){
image_points[0].resize(image_points[1].size());
}
else if (image_points[1].size() > image_points[0].size()){
image_points[1].resize(image_points[0].size());
}
H = findHomography (image_points[0], image_points[1], CV_RANSAC, 3);
cout << "done_matrix" << endl;
}
return H;
}
在调用函数之前,我使用Harris或任何其他检测器检测关键点,并且我传递给函数的图像是由cvtColor函数转换的HSV_image。
错误出现在我之前提到的行中,因为在终端中我可以看到:
1Segmentation Fault: 11
我刚刚完成纠正错误,我认为这是因为我没有有效地使用我的函数。我使用两个"for"语句来交叉检测到的关键点的两个向量,错误有时在第一次出现,而在第二次出现。我真的不知道为什么是错误,我只是改变我的代码做同样的思考更有效,最后它工作。
我只是改变了这几行:
for(vector<KeyPoint>::const_iterator it = keypoints1.begin(); it!= keypoints1.end(); ++it){
// Get the position of left keypoints
float x = (it->pt.x);
float y = (it->pt.y);
cout << "1" << endl;
float saturation_level = chanels_hsv1[1].at<uchar>(x, y);
cout << "2" << endl;
double max_level = 70.0;
cout << "3" << endl;
if ((y < camSize.height/4) && (saturatio_level < max_level) ){
cout << "1:" << endl;
waitKey (100);
cout << "y: " << y;
cout << " Saturation_Level: " << nivel_saturacion << endl;
image_points[0].push_back(Point2f(x,y));
cout << "done" << endl;
cont ++;
}
cont = 0;
for (vector<KeyPoint>::const_iterator it = keypoints2.begin(); it!=keypoints2.end(); ++it) {
// Get the position of left keypoints
float x = (it->pt.x);
float y = (it->pt.y);
float saturation_level = chanels_hsv2[1].at<uchar>(x, y);
double max_level = 70.0;
if ((y < (camSize.height)/4) && (saturation_level < max_level)){
cout << "2" << endl;
waitKey (100);
cout << "y: " << y;
cout << " Saturation_Level: " << nivel_saturacion << endl;
image_points[1].push_back(Point2f(x,y));
cont ++;
}
}
by this:
for (int i = 0; i < good_matches.size(); i++) {
int idx1=good_matches[i].queryIdx;
int idx2=good_matches[i].trainIdx;
if (((keypoints[0][idx1].pt.y < (camSize.height/4)) && (canales_hsv1[1].at<uchar>(keypoints[0][idx1].pt.x, keypoints[0][idx1].pt.y) < nivel_maximo)) || ((keypoints[1][idx2].pt.y < (camSize.height/4)) && (canales_hsv2[1].at<uchar>(keypoints[1][idx1].pt.x, keypoints[1][idx2].pt.y) < nivel_maximo)) ) {
cout << "entro" << endl;
matched_points[0].push_back(keypoints[0][idx1].pt);
matched_points[1].push_back(keypoints[1][idx2].pt);
contador ++;
}
}
目前我只交叉匹配的关键点,而不是所有的关键点,它需要更少的计算机操作,现在它工作正常