使用GDI+进行图像旋转
以前总是用大段大段的cos/sin来写旋转图像的代码，而且旋转出来还经常因为插值的问题导致图像出错。后来稍微研究了一下GDI+，发现直接用GDI+来旋转效果比我自己写好太多了，网上搜到了关于旋转的代码，在这里和大家分享。
/**/
/// <summary>
/// Creates a new Image containing the same image only rotated /// </summary> /// <param name="image">The <see cref="System.Drawing.Image"/> to rotate</param> /// <param name="angle">The amount to rotate the image, clockwise, in degrees</param> /// <returns>A new <see cref="System.Drawing.Bitmap"/> that is just large enough /// to contain the rotated image without cutting any corners off.</returns>
/// <exception cref="System.ArgumentNullException">Thrown if <see cref="image"/> is null.</exception>
public
static
Bitmap RotateImage(Image image,
float
angle)
{ if(image == null) throw new ArgumentNullException("image"); const double pi2 = Math.PI / 2.0; // Why can't C# allow these to be const, or at least readonly // *sigh* I'm starting to talk like Christian Graus :omg: double oldWidth = (double) image.Width; double oldHeight = (double) image.Height; // Convert degrees to radians double theta = ((double) angle) * Math.PI / 180.0; double locked_theta = theta; // Ensure theta is now [0, 2pi) while( locked_theta < 0.0 ) locked_theta += 2 * Math.PI; double newWidth, newHeight; int nWidth, nHeight; // The newWidth/newHeight expressed as ints
Explaination of the calculations#region Explaination of the calculations /**//* * The trig involved in calculating the new width and height * is fairly simple; the hard part was remembering that when * PI/2 <= theta <= PI and 3PI/2 <= theta < 2PI the width and * height are switched. * * When you rotate a rectangle, r, the bounding box surrounding r * contains for right-triangles of empty space. Each of the * triangles hypotenuse's are a known length, either the width or * the height of r. Because we know the length of the hypotenuse * and we have a known angle of rotation, we can use the trig * function identities to find the length of the other two sides. * * sine = opposite/hypotenuse * cosine = adjacent/hypotenuse * * solving for the unknown we get * * opposite = sine * hypotenuse * adjacent = cosine * hypotenuse * * Another interesting point about these triangles is that there * are only two different triangles. The proof for which is easy * to see, but its been too long since I've written a proof that * I can't explain it well enough to want to publish it. * * Just trust me when I say the triangles formed by the lengths * width are always the same (for a given theta) and the same * goes for the height of r. * * Rather than associate the opposite/adjacent sides with the * width and height of the original bitmap, I'll associate them * based on their position. * * adjacent/oppositeTop will refer to the triangles making up the * upper right and lower left corners * * adjacent/oppositeBottom will refer to the triangles making up * the upper left and lower right corners * * The names are based on the right side corners, because thats * where I did my work on paper (the right side). * * Now if you draw this out, you will see that the width of the * bounding box is calculated by adding together adjacentTop and * oppositeBottom while the height is calculate by adding * together adjacentBottom and oppositeTop.
*/ #endregion double adjacentTop, oppositeTop; double adjacentBottom, oppositeBottom; // We need to calculate the sides of the triangles based // on how much rotation is being done to the bitmap. // Refer to the first paragraph in the explaination above for // reasons why. if( (locked_theta >= 0.0 && locked_theta < pi2) || (locked_theta >= Math.PI && locked_theta < (Math.PI + pi2) ) ) { adjacentTop = Math.Abs(Math.Cos(locked_theta)) * oldWidth; oppositeTop = Math.Abs(Math.Sin(locked_theta)) * oldWidth; adjacentBottom = Math.Abs(Math.Cos(locked_theta)) * oldHeight; oppositeBottom = Math.Abs(Math.Sin(locked_theta)) * oldHeight; } else { adjacentTop = Math.Abs(Math.Sin(locked_theta)) * oldHeight; oppositeTop = Math.Abs(Math.Cos(locked_theta)) * oldHeight; adjacentBottom = Math.Abs(Math.Sin(locked_theta)) * oldWidth; oppositeBottom = Math.Abs(Math.Cos(locked_theta)) * oldWidth; } newWidth = adjacentTop + oppositeBottom; newHeight = adjacentBottom + oppositeTop; nWidth = (int) Math.Ceiling(newWidth); nHeight = (int) Math.Ceiling(newHeight); Bitmap rotatedBmp = new Bitmap(nWidth, nHeight); using(Graphics g = Graphics.FromImage(rotatedBmp)) { // This array will be used to pass in the three points that // make up the rotated image Point [] points; /**//* * The values of opposite/adjacentTop/Bottom are referring to * fixed locations instead of in relation to the * rotating image so I need to change which values are used * based on the how much the image is rotating. * * For each point, one of the coordinates will always be 0, * nWidth, or nHeight. This because the Bitmap we are drawing on * is the bounding box for the rotated bitmap. If both of the * corrdinates for any of the given points wasn't in the set above * then the bitmap we are drawing on WOULDN'T be the bounding box * as required. */ if( locked_theta >= 0.0 && locked_theta < pi2 ) { points = new Point[] { new Point( (int) oppositeBottom, 0 ), new Point( nWidth, (int) oppositeTop ), new Point( 0, (int) adjacentBottom ) }; } else if( locked_theta >= pi2 && locked_theta < Math.PI ) { points = new Point[] { new Point( nWidth, (int) oppositeTop ), new Point( (int) adjacentTop, nHeight ), new Point( (int) oppositeBottom, 0 ) }; } else if( locked_theta >= Math.PI && locked_theta < (Math.PI + pi2) ) { points = new Point[] { new Point( (int) adjacentTop, nHeight ), new Point( 0, (int) adjacentBottom ), new Point( nWidth, (int) oppositeTop ) }; } else { points = new Point[] { new Point( 0, (int) adjacentBottom ), new Point( (int) oppositeBottom, 0 ), new Point( (int) adjacentTop, nHeight ) }; } g.DrawImage(image, points); } return rotatedBmp; }