VC 图片旋转难题 ！ 求高手指点（还可加分）！！
用StretchBlt函数实现将图片旋转90度和270度，请高手指点，详细说明！
如果有其它函数也可；请详细说明一下，谢谢！
[解决办法]
旋转得自己做把
[解决办法]
// RotateDIB - Create a new bitmap with rotated image// Returns - Returns new bitmap with rotated image// hDIB - Device-independent bitmap to rotate// fDegrees - Angle of rotation in degree// clrBack - Color of pixels in the resulting bitmap that do// not get covered by source pixelsHDIB RotateDIB(HDIB hDIB, double fDegrees, COLORREF clrBack){ WaitCursorBegin(); // Get source bitmap ec
[解决办法]
这里有代码下载。
[解决办法]
函数: TransparentBit 解释: BOOL TransparentBit(HDC hdc, //目标DCint nXOriginDest, //目标X偏移int nYOriginDeST, //目标Y偏移int nWidthDest, //目标宽度int nHeightDest, //目标高度HDC hdcSrc, //源DCint nXOriginSrc, //源X起点int nYOriginSrc, //源Y起点int nWidthSrc, //源高度int nHeightSrc, //源高度UINT crTransparent //透明色,COLORREF类型)函数: LoadImage 解释: 从磁盘文件或者可执行文件资源中加载一个Windows位图 函数: LoadImage 解释: HANDLE LoadImage(HINSTANCE hInstance, LPCTSTR lpszName,UINT uType,int cxDesired,int cyDesired,UINT fuLoad);
[解决办法]
漏了链接地址了
http://www.codeproject.com/KB/graphics/rotatebyshear.aspx
[解决办法]
/// <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;
}
Graphics* g;GraphicsPath myGraphicsPath(FillModeAlternate); int offsetrx=m_EllipseData.point1.x -origin.X; int offsetry=m_EllipseData.point1 .y -origin.Y;RectF myRectangle(offsetrx,offsetry ,m_EllipseData.point2.x-m_EllipseData.point1.x,m_EllipseData.point2.y-m_EllipseData.point1 .y ); myGraphicsPath.AddEllipse (myRectangle); Matrix* myPathMatrix=new Matrix(); myPathMatrix->Rotate(nRotateAngle, MatrixOrderAppend); myPathMatrix->Translate(origin.X,origin.Y,MatrixOrderAppend ); myGraphicsPath.Transform(myPathMatrix); g->FillPath(&myBrush, &myGraphicsPath);