GraphicsPath 和 OutOfMemoryException

Question

我有以下情况

private bool IsPathVisible(Rectangle detectorRectangle, GraphicsPath path, Pen pen)
{
    path.Widen(pen);
    return IsPathVisible(detectorRectangle, path);
}

当 path 点是同一点时，我收到 OutOfMemoryException （使用 Widen 函数）。

我该如何管理它？

Answer 1

这是笔和加宽方法的错误。确保路径的起点和路径的终点不相同。

这是一个演示：

private void panel1_Paint(object sender, PaintEventArgs e)
{
  //This works:
  using (GraphicsPath path = new GraphicsPath())
  {
    path.AddLine(new Point(16, 16), new Point(20, 20));
    path.Widen(Pens.Black);
    e.Graphics.DrawPath(Pens.Black, path);
  }

  //This does not:
  using (GraphicsPath path = new GraphicsPath())
  {
    path.AddLine(new Point(20, 20), new Point(20, 20));
    path.Widen(Pens.Black);
    e.Graphics.DrawPath(Pens.Black, path);
  }
}

这是向 Microsoft 报告的位置： GraphicsPath.Widen 如果路径有一个点，则抛出 OutOfMemoryException

Answer 2

我也曾遭受过这种异常的困扰。建议如下：

1. 在扩大之前保存点，以查看导致 OutOfMemoryException 的确切点：

   private bool IsPathVisible(矩形检测器矩形, GraphicsPath 路径, Pen pen)
   {
       var points = path.PathPoints.Clone() as PointF[];
       路径.加宽(笔);
       返回 IsPathVisible(探测器矩形, 路径);
   }
   

您可能会看到可能存在具有相同坐标的后续点。他们实际上造成了问题。

2. 此外，GraphicsPath 可以由多个子路径组成。为了进行可靠的命中测试，我建议执行以下操作：

   public Region[] CreateRegionFromGraphicsPath(GraphicsPath 路径，Pen 加宽Pen)
   {
       var Regions = new List();
   
       var itPath = new GraphicsPathIterator(路径);
       itPath.Rewind();
       var curSubPath = new GraphicsPath();
   
       for (int i = 0; i < itPath.SubpathCount; i++)
       {
           bool 已关闭；
           itPath.NextSubpath(curSubPath, out isClosed);
   
           if (!isClosed && CanWiden(curSubPath)) curSubPath.Widen(wideningPen); // 加宽非闭合路径
   
           int 区域索引 = i / 100; // 最大区域扫描矩形数
   
           if (regions.Count 
   /// 确定加宽此图形路径是否不会导致异常
   /// 
   公共静态 bool CanWiden(GraphicsPath gp)
   {
       const floatregionPointsTolerance = 1e-8f;
       var pts = gp.PathPoints;
       if (pts.Length < 2) 返回 false;
       for (int i = 1; i < pts.Length; i++)
       {
           if (Math.Abs​​(pts[i-1].X - pts[i].X) 

然后您只需为区域调用 IsVisible 来查找是否有任何区域被命中

Answer 3

如果路径 IsPoint，则不进行加宽操作。

<System.Runtime.CompilerServices.Extension()> _
Public Function IsPoint(ByVal path As System.Drawing.Drawing2D.GraphicsPath) As Boolean
    If path Is Nothing Then Throw New ArgumentNullException("path")

    If path.PathPoints.Count < 2 Then Return True

    If path.PathPoints(0) <> path.PathPoints(path.PathPoints.Count - 1) Then Return False

    For i = 1 To path.PathPoints.Count - 1
     If path.PathPoints(i - 1) <> path.PathPoints(i) Then Return False
    Next i

    ' if all the points are the same
    Return True
End Function 

Answer 4

以下代码会导致 .Net 4.0（以及可能更高版本）中的 DrawPath 出现 OutOfMemory。
我设法使用 LineCap.Flat 而不是 LineCap.NoAnchor 绕过它：

public void TestDrawPath()
{
    PointF[] points = new PointF[13]
    {
        new PointF(0.491141558f, 1.53909028f),
        new PointF(0.491141558f, 1.55148673f),
        new PointF(0.4808829f, 1.56153619f),
        new PointF(0.468228281f, 1.56153619f),
        new PointF(0.4555736f, 1.56153619f),
        new PointF(0.445314974f, 1.55148673f),
        new PointF(0.445314974f, 1.53909028f),
        new PointF(0.445314974f, 1.52669382f),
        new PointF(0.4555736f, 1.51664436f),
        new PointF(0.468228281f, 1.51664436f),
        new PointF(0.4808829f, 1.51664436f),
        new PointF(0.491141558f, 1.52669382f),
        new PointF(0.491141558f, 1.53909028f)
    };
    byte[] types = new byte[13] { 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 131 };

    using (Bitmap bitmap = new Bitmap(2, 2))
    using (Graphics g = Graphics.FromImage(bitmap))
    {
        using (Pen pen = new Pen(Color.Black))
        using (GraphicsPath path = new GraphicsPath(points, types))
        {
            pen.StartCap = LineCap.NoAnchor;
            pen.EndCap = LineCap.NoAnchor;
            g.DrawPath(pen, path);
        }
    }
}