我不确定您是否可以一次获得一个轮廓。但如果你有一个vector<vector<Point> >
,你可以迭代每个轮廓,如下所示:
using namespace std;
vector<vector<Point> > contours;
// use findContours or other function to populate
for(size_t i=0; i<contours.size(); i++) {
// use contours[i] for the current contour
for(size_t j=0; j<contours[i].size(); j++) {
// use contours[i][j] for current point
}
}
// Or use an iterator
vector<vector<Point> >::iterator contour = contours.begin(); // const_iterator if you do not plan on modifying the contour
for(; contour != contours.end(); ++contour) {
// use *contour for current contour
vector<Point>::iterator point = contour->begin(); // again, use const_iterator if you do not plan on modifying the contour
for(; point != contour->end(); ++point) {
// use *point for current point
}
}
因此,为了更好地回答您关于h_next
. 给定迭代器 ,it
在 中vector
,下一个元素将是it+1
。示例用法:
vector<vector<Point> >::iterator contour = contours.begin();
vector<vector<Point> >::iterator next = contour+1; // behavior like h_next