.net - Region.IsVisible() 不按文档工作 (?) [GDI+]

Question

我看到 Region.IsVisible(rectangle) 没有按我预期的那样工作。
所以，是我谁期望不应该，或者是方法那不是应该的吗？？!

我有以下情况:

alt text http://lh4.ggpht.com/_1TPOP7DzY1E/TCmxn6Tzn2I/AAAAAAAADRc/GJhbStCvabQ/s800/Capture3.gif alt text http://lh5.ggpht.com/_1TPOP7DzY1E/TCmuVyrgpTI/AAAAAAAADRU/yLNn-jZQDNA/s800/Capture2.gif

以及以下代码:

private void Form1_Paint(object sender, PaintEventArgs e)
{
    Point[] points1 = new Point[] {
        new Point(50, 30),
        new Point(70, 30),
        new Point(40, 40),
        new Point(60, 70),
        new Point(30, 50)
    };

    Point[] points2 = new Point[] {
        new Point(70, 150),
        new Point(50, 110 ),
        new Point(60, 80),
        new Point(90, 80),
        new Point(140, 60)                
    };

    Point[] points3 = new Point[] {
        new Point(100, 10),
        new Point(130, 40)
    };

    GraphicsPath path1 = new GraphicsPath();
    GraphicsPath path2 = new GraphicsPath();
    GraphicsPath path3 = new GraphicsPath();

    path1.AddLines(points1);
    path2.AddLines(points2);
    path3.AddLines(points3);

    e.Graphics.DrawPath(Pens.DarkBlue, path1);
    e.Graphics.DrawPath(Pens.DarkGreen, path2);
    e.Graphics.DrawPath(Pens.DarkOrange, path3);

    Region r1 = new Region(path1);
    Region r2 = new Region(path2);
    Region r3 = new Region(path3);

    // Create the first rectangle and draw it to the screen in blue.
    Rectangle blueRect = new Rectangle(20, 20, 100, 100);
    e.Graphics.DrawRectangle(Pens.Blue, blueRect);

    bool contained;

    // Display the result.                        
    ControlPaint.DrawGrid(e.Graphics, this.ClientRectangle,
        new Size(10, 10), Color.Red);

    contained = r1.IsVisible(blueRect);
    e.Graphics.DrawString("Path blue contained = " + contained.ToString(),
        Font, myBrush, new PointF(20, 160));

    contained = r2.IsVisible(blueRect);
    e.Graphics.DrawString("Path green contained = " + contained.ToString(),
        Font, Brushes.Black, new PointF(20, 180));

    contained = r3.IsVisible(blueRect);
    e.Graphics.DrawString("Path orange contained = " + contained.ToString(),
        Font, Brushes.Black, new PointF(20, 200));
}

此外，不在该区域内的路径可能是“可见的”:

Point[] points3 = new Point[] {
  new Point(15, 35),
  new Point(15, 130),
  new Point(60 ,130)
};

编辑:
甚至 Intersect 对第二个 L 路径不起作用:

Point[] points3 = new Point[] {
    new Point(10, 40),
    new Point(10, 130),
    new Point(50 ,130)
};

r3.Intersect(blueRect);
bool contained = !(r1.IsEmpty(e.Graphics)); 
e.Graphics.DrawString("Path orange contained = " + contained.ToString(),
    Font, Brushes.Black, new PointF(20, 200)); // TRUE! instead of desired FALSE

Answer 1

从橙色的 L 形示例中，我看到您误解了您的代码的作用。

从路径构建区域不会产生绘制路径形式的区域。从 L 形橙色路径构建区域不会产生 L 形单像素宽度区域。路径通过连接两端形成三角形而闭合。该区域是该三角形的内部，蓝色矩形显然部分包含在该区域中。

对于单条橙色线的初始示例，结果区域是只有两个角的退化多边形 - 仍然看起来像一条线，并且在与该线正交的方向上具有零宽度。因此，该区域的面积为零且不包含任何内容(如果区域是闭集，则该区域的边界上的点可能除外)。

您真正想要做的是针对蓝色矩形的矩形区域执行路径的可见性测试。据我所知，这样做没有内置支持。