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C++ PathVector::back方法代码示例

本文整理汇总了C++中geom::PathVector::back方法的典型用法代码示例。如果您正苦于以下问题:C++ PathVector::back方法的具体用法?C++ PathVector::back怎么用?C++ PathVector::back使用的例子?那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在geom::PathVector的用法示例。


在下文中一共展示了PathVector::back方法的2个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C++代码示例。

示例1: l

/*
 * Converts all segments in all paths to Geom::LineSegment or Geom::HLineSegment or
 * Geom::VLineSegment or Geom::CubicBezier.
 */
Geom::PathVector
pathv_to_linear_and_cubic_beziers( Geom::PathVector const &pathv )
{
    Geom::PathVector output;

    for (Geom::PathVector::const_iterator pit = pathv.begin(); pit != pathv.end(); ++pit) {
        output.push_back( Geom::Path() );
        output.back().start( pit->initialPoint() );
        output.back().close( pit->closed() );

        for (Geom::Path::const_iterator cit = pit->begin(); cit != pit->end_open(); ++cit) {
            if (is_straight_curve(*cit)) {
                Geom::LineSegment l(cit->initialPoint(), cit->finalPoint());
                output.back().append(l);
            } else {
                Geom::BezierCurve const *curve = dynamic_cast<Geom::BezierCurve const *>(&*cit);
                if (curve && curve->order() == 3) {
                    Geom::CubicBezier b((*curve)[0], (*curve)[1], (*curve)[2], (*curve)[3]);
                    output.back().append(b);
                } else {
                    // convert all other curve types to cubicbeziers
                    Geom::Path cubicbezier_path = Geom::cubicbezierpath_from_sbasis(cit->toSBasis(), 0.1);
                    output.back().append(cubicbezier_path);
                }
            }
        }
    }
    
    return output;
}
开发者ID:asitti,项目名称:inkscape,代码行数:34,代码来源:geom.cpp

示例2: ls

/*
 * Converts all segments in all paths to Geom::LineSegment.  There is an intermediate
 * stage where some may be converted to beziers.  maxdisp is the maximum displacement from
 * the line segment to the bezier curve; ** maxdisp is not used at this moment **.
 *
 * This is NOT a terribly fast method, but it should give a solution close to the one with the
 * fewest points.
 */
Geom::PathVector
pathv_to_linear( Geom::PathVector const &pathv, double /*maxdisp*/)
{
    Geom::PathVector output;
    Geom::PathVector tmppath = pathv_to_linear_and_cubic_beziers(pathv);
    
    // Now all path segments are either already lines, or they are beziers.

    for (Geom::PathVector::const_iterator pit = tmppath.begin(); pit != tmppath.end(); ++pit) {
        output.push_back( Geom::Path() );
        output.back().start( pit->initialPoint() );
        output.back().close( pit->closed() );

        for (Geom::Path::const_iterator cit = pit->begin(); cit != pit->end_open(); ++cit) {
            if (is_straight_curve(*cit)) {
                Geom::LineSegment ls(cit->initialPoint(), cit->finalPoint());
                output.back().append(ls);
            } 
            else { /* all others must be Bezier curves */
                Geom::BezierCurve const *curve = dynamic_cast<Geom::BezierCurve const *>(&*cit);
                Geom::CubicBezier b((*curve)[0], (*curve)[1], (*curve)[2], (*curve)[3]);
                std::vector<Geom::Point> bzrpoints = b.points();
                Geom::Point A = bzrpoints[0];
                Geom::Point B = bzrpoints[1];
                Geom::Point C = bzrpoints[2];
                Geom::Point D = bzrpoints[3];
                std::vector<Geom::Point> pointlist;
                pointlist.push_back(A);
                recursive_bezier4(
                   A[X], A[Y], 
                   B[X], B[Y], 
                   C[X], C[Y], 
                   D[X], D[Y],
                   pointlist, 
                   0);
                pointlist.push_back(D);
                Geom::Point r1 = pointlist[0];
                for (unsigned int i=1; i<pointlist.size();i++){
                   Geom::Point prev_r1 = r1;
                   r1 = pointlist[i];
                   Geom::LineSegment ls(prev_r1, r1);
                   output.back().append(ls);
                }
                pointlist.clear();
           }
        }
    }
    
    return output;
}
开发者ID:asitti,项目名称:inkscape,代码行数:58,代码来源:geom.cpp


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