C# 获取RotateTransform的结果大小
我有以下代码用于在C中旋转图像:C# 获取RotateTransform的结果大小,c#,winforms,rotation,system.drawing,C#,Winforms,Rotation,System.drawing,我有以下代码用于在C中旋转图像: private Bitmap RotateImage(Bitmap b, float angle) { //create a new empty bitmap to hold rotated image Bitmap returnBitmap = new Bitmap(b.Width, b.Height); //make a graphics object from the empty bitmap Graphics g = G
private Bitmap RotateImage(Bitmap b, float angle)
{
//create a new empty bitmap to hold rotated image
Bitmap returnBitmap = new Bitmap(b.Width, b.Height);
//make a graphics object from the empty bitmap
Graphics g = Graphics.FromImage(returnBitmap);
//move rotation point to center of image
g.TranslateTransform((float)returnBitmap.Width / 2, (float)returnBitmap.Height / 2);
//rotate
g.RotateTransform(angle);
//move image back
g.TranslateTransform(-(float)b.Width / 2, -(float)b.Height / 2);
//draw passed in image onto graphics object
g.DrawImage(b, new Rectangle(new Point(0, 0), new Size(b.Width, b.Height)));
return returnBitmap;
}
它工作得非常好,只是当结果超出原始边界时会进行剪辑
据我所知,我必须将returnBitmap的大小设置为旋转后图像的大小。但是,如何找到结果的大小,从而相应地设置新位图的大小?毕达哥拉斯。它在90/270角度下从原始到sqrtw^2+h^2的任何位置。我敢打赌它是由sine max在90确定的。您需要旋转原始图像的四个角,并计算新坐标的边界框:
private static Bitmap RotateImage(Image b, float angle)
{
var corners = new[]
{new PointF(0, 0), new Point(b.Width, 0), new PointF(0, b.Height), new PointF(b.Width, b.Height)};
var xc = corners.Select(p => Rotate(p, angle).X);
var yc = corners.Select(p => Rotate(p, angle).Y);
//create a new empty bitmap to hold rotated image
Bitmap returnBitmap = new Bitmap((int)Math.Abs(xc.Max() - xc.Min()), (int)Math.Abs(yc.Max() - yc.Min()));
...
}
/// <summary>
/// Rotates a point around the origin (0,0)
/// </summary>
private static PointF Rotate(PointF p, float angle)
{
// convert from angle to radians
var theta = Math.PI*angle/180;
return new PointF(
(float) (Math.Cos(theta)*(p.X) - Math.Sin(theta)*(p.Y)),
(float) (Math.Sin(theta)*(p.X) + Math.Cos(theta)*(p.Y)));
}
看看我的评论