我在编写 beizer 函数时遇到问题。我在 c# 中编写了一个简单的代码:
public static PointF[] BeizerFunction (int interval, PointF point0, PointF point1, PointF point2) {
//x = (1 - t) * (1 - t) * p[0].x + 2 * (1 - t) * t * p[1].x + t * t * p[2].x;
//y = (1 - t) * (1 - t) * p[0].y + 2 * (1 - t) * t * p[1].y + t * t * p[2].y;
var Points = new PointF[interval];
var time = 1.0f / (float) interval;
for (var i=0; i<Points.Length; i++) {
var point = Points [i];
point = new PointF ();
point.X = (1 - time) * (1 - time) * point0.X
+ 2 * (1 - time) * time * point1.X
+ time * time * point2.X;
point.Y = (1 - time) * (1 - time) * point0.Y
+ 2 * (1 - time) * time * point1.Y
+ time * time * point2.Y;
Points [i] = point;
time ++;
}
return Points;
}
通过应用程序听鼠标移动,所以我猜鼠标指针将是这个函数的控制点。如果正确,上面的代码应该给我一个来自三个点的贝泽曲线。
实际上,有多条曲线,所以会有超过 3 个点。为了将 beize 曲线连接在一起,我做了这样的事情。
Bezier(p0.5, p1, p1.5);
Bezier(p1.5, p2, p2.5);
Bezier(p2.5, p3, p3.5);
仅绘制 3 个点的 c# 代码是这样的
var p0 = new PointF (50, 50);
var p1 = new PointF (100, 100);
var p2 = new PointF (150, 50);
var points = QuadraticBezierFunction.BeizerFunction (100, p0, p1, p2);
for (var i=0; points != null && i<points.Length-1; i=i+1)
canvas.DrawLine (points[i].X, points[i].Y, points[i+1].X, points[i+1].Y, new Android.Graphics.Paint ());
当我尝试绘制曲线时,它看起来不是曲线。