在 C# 中旋转图像时如何防止剪切？

Question

我刚刚经历了很多事情，试图找出如何让图像旋转。这可行，但现在它正在剪辑，我不太确定如何让它停止......我正在使用这个rotateImage方法：

public static Image RotateImage(Image img, float rotationAngle)
    {
        //create an empty Bitmap image
        Bitmap bmp = new Bitmap(img.Width, img.Height);

        //turn the Bitmap into a Graphics object
        Graphics gfx = Graphics.FromImage(bmp);

        //now we set the rotation point to the center of our image
        gfx.TranslateTransform((float)bmp.Width / 2, (float)bmp.Height / 2);

        //now rotate the image
        gfx.RotateTransform(rotationAngle);

        gfx.TranslateTransform(-(float)bmp.Width / 2, -(float)bmp.Height / 2);

        //set the InterpolationMode to HighQualityBicubic so to ensure a high
        //quality image once it is transformed to the specified size
        gfx.InterpolationMode = InterpolationMode.HighQualityBicubic;

        //now draw our new image onto the graphics object
        gfx.DrawImage(img, new System.Drawing.Point(0, 0));

        //dispose of our Graphics object
        gfx.Dispose();

        //return the image
        return bmp;
    }

我尝试使空位图变大，但这仅适用于一侧，因为图像固定到位图的左上角。任何想法，将不胜感激！

Answer 1

我发现从其他网站一些帮助。下面是我落得这样做对于那些你想知道谁：

// Rotates the input image by theta degrees around center.
    public static Bitmap RotateImage(Bitmap bmpSrc, float theta)
    {
        Matrix mRotate = new Matrix();
        mRotate.Translate(bmpSrc.Width / -2, bmpSrc.Height / -2, MatrixOrder.Append);
        mRotate.RotateAt(theta, new System.Drawing.Point(0, 0), MatrixOrder.Append);
        using (GraphicsPath gp = new GraphicsPath())
        {  // transform image points by rotation matrix
            gp.AddPolygon(new System.Drawing.Point[] { new System.Drawing.Point(0, 0), new System.Drawing.Point(bmpSrc.Width, 0), new System.Drawing.Point(0, bmpSrc.Height) });
            gp.Transform(mRotate);
            System.Drawing.PointF[] pts = gp.PathPoints;

            // create destination bitmap sized to contain rotated source image
            Rectangle bbox = boundingBox(bmpSrc, mRotate);
            Bitmap bmpDest = new Bitmap(bbox.Width, bbox.Height);

            using (Graphics gDest = Graphics.FromImage(bmpDest))
            {  // draw source into dest
                Matrix mDest = new Matrix();
                mDest.Translate(bmpDest.Width / 2, bmpDest.Height / 2, MatrixOrder.Append);
                gDest.Transform = mDest;
                gDest.DrawImage(bmpSrc, pts);
                return bmpDest;
            }
        }
    }

    private static Rectangle boundingBox(Image img, Matrix matrix)
    {
        GraphicsUnit gu = new GraphicsUnit();
        Rectangle rImg = Rectangle.Round(img.GetBounds(ref gu));

        // Transform the four points of the image, to get the resized bounding box.
        System.Drawing.Point topLeft = new System.Drawing.Point(rImg.Left, rImg.Top);
        System.Drawing.Point topRight = new System.Drawing.Point(rImg.Right, rImg.Top);
        System.Drawing.Point bottomRight = new System.Drawing.Point(rImg.Right, rImg.Bottom);
        System.Drawing.Point bottomLeft = new System.Drawing.Point(rImg.Left, rImg.Bottom);
        System.Drawing.Point[] points = new System.Drawing.Point[] { topLeft, topRight, bottomRight, bottomLeft };
        GraphicsPath gp = new GraphicsPath(points,
                                                            new byte[] { (byte)PathPointType.Start, (byte)PathPointType.Line, (byte)PathPointType.Line, (byte)PathPointType.Line });
        gp.Transform(matrix);
        return Rectangle.Round(gp.GetBounds());
    }

Answer 2

按着这些次序：

1. 创建空的目标图像，其宽度和高度 = 旋转图像的边界框
2. 将源图像绘制到目标图像上。源的 (0,0) 将映射到目标的 (ox,oy)
3. 现在执行旋转目标图像的步骤。

完成上述步骤的详细信息：

W,H = 目的地的宽度和高度

w,h = 源的宽度和高度

c=|cos(θ)|

s=|sin(θ)|

w * c + h * s = W

w * s + h * c = H

ox = ( W - w ) / 2

oy = ( H - h ) / 2

Answer 3

public Bitmap rotateImage(Bitmap b, float angle)
    {
        if (angle > 0)
        {
            int l = b.Width;
            int h = b.Height;
            double an = angle * Math.PI / 180;
            double cos = Math.Abs(Math.Cos(an));
            double sin = Math.Abs(Math.Sin(an));
            int nl = (int)(l * cos + h * sin);
            int nh = (int)(l * sin + h * cos);
            Bitmap returnBitmap = new Bitmap(nl, nh);
            Graphics g = Graphics.FromImage(returnBitmap);
            g.TranslateTransform((float)(nl-l) / 2, (float)(nh-h) / 2);
            g.TranslateTransform((float)b.Width / 2, (float)b.Height / 2);
            g.RotateTransform(angle);
            g.TranslateTransform(-(float)b.Width / 2, -(float)b.Height / 2);
            g.DrawImage(b, new Point(0, 0));
            return returnBitmap;
        }
        else return b;
    }

Answer 4

一个旋转的图像可能需要较大的位图被包含不削波。一个相当简单的方法来计算边界是变换矩阵向原边框的角落。由于新的位图较大时，原始图像必须被绘制为使得它的中心，而不是在（0,0）。

这证明在以下修订您的代码：

public static Image RotateImage(Image img, float rotationAngle)
{
    int minx = int.MaxValue, maxx = int.MinValue, miny = int.MaxValue, maxy = int.MinValue;

    using (Bitmap bmp = new Bitmap(1, 1)) // Dummy bitmap, so we can use TransformPoints to figure out the correct size.
    {
        using (Graphics g = Graphics.FromImage(bmp))
        {
            g.TranslateTransform((float)img.Width / 2, (float)img.Height / 2);
            g.RotateTransform(rotationAngle);
            g.TranslateTransform(-(float)img.Width / 2, -(float)img.Height / 2);

            Point[] pts = new Point[4];
            pts[0] = new Point(0, 0);
            pts[1] = new Point(img.Width, 0);
            pts[2] = new Point(img.Width, img.Height);
            pts[3] = new Point(0, img.Height);
            g.TransformPoints(CoordinateSpace.Device, CoordinateSpace.World, pts);

            foreach (Point pt in pts)
            {
                minx = Math.Min(minx, pt.X);
                maxx = Math.Max(maxx, pt.X);
                miny = Math.Min(miny, pt.Y);
                maxy = Math.Max(maxy, pt.Y);
            }
        }
    }

    Bitmap bmp2 = new Bitmap(maxx - minx, maxy - miny);
    using (Graphics g = Graphics.FromImage(bmp2))
    {
        g.TranslateTransform((float)bmp2.Width / 2, (float)bmp2.Height / 2);
        g.RotateTransform(rotationAngle);
        g.TranslateTransform(-(float)bmp2.Width / 2, -(float)bmp2.Height / 2);
        g.InterpolationMode = InterpolationMode.HighQualityBicubic;
        g.DrawImage(img, bmp2.Width / 2 - img.Width / 2, bmp2.Height / 2 - img.Height / 2);
    }

    return bmp2;
}