CVPolylines如何工作

Question

遵循在学习OpenCV中给出的Delaunay三角剖分之后，我很难理解此片段，这是负责绘制镶嵌的最终作品，此处draw_subdiv_facet正在为一个voroni送出一个voroni时间

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 );
}

这是负责在多边形中绘制线条和着色的原因

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

这些点在-1000或更大的范围内，因此如何仍在绘制点。

Answer 1

目前尚不清楚您要问什么。如果点"按1000或更多的顺序"，那么源图像可能很大。这些点相对于源图像，而不是窗口。如果需要将点放入绘图窗口中，则需要手动缩放点。

Answer 2

你是对的。它绘制了可能在200多个以上的坐标中的10分中绘制了10分，顺便说一句，我只是没有看到这些点，而且很困惑，但它们一直在那里。谢谢。