解析在Direct2D中畫Bezier曲線的實現方法

2020-01-26 16:11:50

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Direct2D通過ID2D1RenderTarget接口支持基本圖元（直線，矩形，圓角矩形，橢圓等）的繪制，然而，此接口并未提供對曲線繪制的直接支持。因此，想要使用Direct2D繪制一段通過指定點的曲線，比如Bezier曲線，必須借助于DrawGeometry()方法間接實現。需要通過一定的算法，將指定點轉換為定義Path的控制點。幸運的是，codproject上已經有人做了這項工作，給出了相應的轉換算法，并給出了C#版的實現：
Draw a Smooth Curve through a Set of 2D Points with Bezier Primitives
C#的代碼可以很容易的轉換成C++版本的，下面是我轉換的一個用于Direct2D的繪制Bezier曲線的C++函數：

復制代碼代碼如下:

/// <summary>
/// Refer to : http://www.codeproject.com/KB/graphics/BezierSpline.aspx
/// Solves a tridiagonal system for one of coordinates (x or y) of first Bezier control points.
/// </summary>
/// <param name="rhs">Right hand side vector.</param>
/// <param name="x">Solution vector.</param>
void GetFirstControlPoints(
     __in const std::vector<FLOAT>& rhs,
     __out std::vector<FLOAT>& x )
{
     ATLASSERT(rhs.size()==x.size());
     int n = rhs.size();
     std::vector<FLOAT> tmp(n);    // Temp workspace.

     FLOAT b = 2.0f;
     x[0] = rhs[0] / b;
     for (int i = 1; i < n; i++) // Decomposition and forward substitution.
     {
         tmp[i] = 1 / b;
         b = (i < n-1 ? 4.0f : 3.5f) - tmp[i];
         x[i] = (rhs[i] - x[i-1]) / b;
     }
     for (int i = 1; i < n; i++)
     {
         x[n-i-1] -= tmp[n-i] * x[n-i]; // Back substitution.
     }
}

/// <summary>
/// Refer to : http://www.codeproject.com/KB/graphics/BezierSpline.aspx
/// Get open-ended Bezier Spline Control Points.
/// </summary>
/// <param name="knots">Input Knot Bezier spline points.</param>
/// <param name="firstCtrlPt">Output First Control points array of knots.size()-1 length.</param>
/// <param name="secondCtrlPt">Output Second Control points array of knots.size()-1 length.</param>
void GetCurveControlPoints(
     __in const std::vector<D2D1_POINT_2F>& knots,
     __out std::vector<D2D1_POINT_2F>& firstCtrlPt,
     __out std::vector<D2D1_POINT_2F>& secondCtrlPt )
{
     ATLASSERT( (firstCtrlPt.size()==secondCtrlPt.size())
         && (knots.size()==firstCtrlPt.size()+1) );

     int n = knots.size()-1;
     ATLASSERT(n>=1);

     if (n == 1)
     {
         // Special case: Bezier curve should be a straight line.
         // 3P1 = 2P0 + P3
         firstCtrlPt[0].x = (2 * knots[0].x + knots[1].x) / 3.0f;
         firstCtrlPt[0].y = (2 * knots[0].y + knots[1].y) / 3.0f;

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