Question

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 = 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;
}

它非常有效，只是当它超过原始边界时会剪切结果。

据我所知，我必须将returnBitmap的大小设置为旋转后的图像大小。但是，如何确定结果的大小，相应地设置新位图的大小？

Answer 1

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)));
    }

Answer 2

毕达哥拉斯。它是从原始到sqrt（w ^ 2 + h ^ 2）在90/270角度的任何地方。我打赌它是由正弦决定的（最大值为90）。

获得RotateTransform的结果大小

2 个答案: