Question

在学习opencv后给出的delaunay三角剖分之后，我在理解这个片段时遇到了一些麻烦，这是最后一块负责绘制图形的片段，这里draw_subdiv_facet一次被送入一个voroni边缘

static void draw_subdiv_facet( IplImage* img, CvSubdiv2DEdge edge )
{
    CvSubdiv2DEdge t = edge;
    int i, count = 0;

    //cvpoint structure
    //param x:  x-coordinate of the point.
    //param y:  y-coordinate of the point.
    //param point:  the point to convert.
    CvPoint* buf = 0;

    // count number of edges in facet
    do
    {
        count++;
        t = cvSubdiv2DGetEdge( t, CV_NEXT_AROUND_LEFT );
    } while (t != edge );
    cout<<"\ncount is : "<<count<<endl;

    //allocate the array
    buf = (CvPoint*)malloc( count * sizeof(buf[0]));

    // gather points
    t = edge;
    for( i = 0; i < count; i++ )
    {
        //
        CvSubdiv2DPoint* pt = cvSubdiv2DEdgeOrg( t );          
        if( !pt ) break;


        buf[i] = cvPoint( cvRound(pt->pt.x), cvRound(pt->pt.y));
        cout<<"pt.x is : "<<cvRound(pt->pt.x);
        cout<<"   pt.y is : "<<cvRound(pt->pt.y)<<endl;
        cout<<"converted to cvPoint gives"<<buf[i].x<<" , "<<buf[i].y<<endl;
        t = cvSubdiv2DGetEdge( t, CV_NEXT_AROUND_LEFT );
    }

    if( i == count )   
    {
        CvSubdiv2DPoint* pt = cvSubdiv2DEdgeDst( cvSubdiv2DRotateEdge( edge, 1 ));
        //cvFillConvexPoly( img, buf, count, CV_RGB(rand()&255,rand()&255,rand()&255), CV_AA, 0 );


        CvPoint xx = buf[0];
        cout<<"located at "<<xx.x<<","<<xx.y<<endl;
        cvPolyLine( img, &buf, &count, 1, 1, CV_RGB(0,0,0), 1, CV_AA, 0);
        draw_subdiv_point( img, pt->pt, CV_RGB(0,0,0));
    }
    free( buf );
}

如你所见，它负责绘制多边形中的线条和着色，但是cout输出的点远大于窗口本身，即画布

CvRect rect = { 0, 0, 600, 600 };
img = cvCreateImage( cvSize(rect.width,rect.height), 8, 3 );

这些点大约是-1000或更多，所以它仍然是如何绘制点。

Answer 1

目前还不清楚你在问什么。如果这些点是“大约1000或更多”，则源图像可能很大。这些点是相对于源图像而不是窗口。如果需要它们适合绘图窗口，您需要自己手动缩放点。

Answer 2

你是对的，我的错。它绘制了200多个点中的10个，坐标为btw 0和1000，我只是没有看到这些点而感到困惑，但他们一直都在那里。感谢。

cvpolylines如何运作

2 个答案: