WIN32编程中,怎么实现位图的旋转???

Top0ne 2006-05-09 05:42:34
请问高手们:
WIN32编程中.怎么才能让位图旋转任意角度呀?
...全文
664 13 打赏 收藏 转发到动态 举报
写回复
用AI写文章
13 条回复
切换为时间正序
请发表友善的回复…
发表回复
doudou52520 2007-04-08
  • 打赏
  • 举报
回复
mark
e_feeling 2007-03-21
  • 打赏
  • 举报
回复
gz
cuterain 2006-10-30
  • 打赏
  • 举报
回复
mark
CUG122032 2006-05-09
  • 打赏
  • 举报
回复
GDI+
CUG122032 2006-05-09
  • 打赏
  • 举报
回复
GDI+
Kudeet 2006-05-09
  • 打赏
  • 举报
回复
Rotate a Bitmap at Any Angle Without GetPixel/SetPixel
http://www.codeguru.com/cpp/g-m/gdi/article.php/c3693/
Kudeet 2006-05-09
  • 打赏
  • 举报
回复
http://www.codeguru.com/cpp/g-m/bitmap/specialeffects/article.php/c1743/
介绍的3种方法
lyl_rabbit 2006-05-09
  • 打赏
  • 举报
回复
建议使用CxImage类来处理

http://www.codeproject.com/bitmap/cximage.asp
iamwjp 2006-05-09
  • 打赏
  • 举报
回复
函数三: GetRotatedBitmap()使用DIB

// GetRotatedBitmap - Create a new bitmap with rotated image
// Returns - Returns new bitmap with rotated image
// hDIB - Device-independent bitmap to rotate
// radians - Angle of rotation in radians
// clrBack - Color of pixels in the resulting bitmap that do
// not get covered by source pixels
HANDLE GetRotatedBitmap( HANDLE hDIB, float radians, COLORREF clrBack)
{
// Get source bitmap info
BITMAPINFO &bmInfo = *(LPBITMAPINFO)hDIB ;
int bpp = bmInfo.bmiHeader.biBitCount; // Bits per pixel

int nColors = bmInfo.bmiHeader.biClrUsed ? bmInfo.bmiHeader.biClrUsed
:
1 << bpp;
int nWidth = bmInfo.bmiHeader.biWidth;
int nHeight = bmInfo.bmiHeader.biHeight;
int nRowBytes = ((((nWidth * bpp) + 31) & ~31) / 8);

// Make sure height is positive and biCompression is BI_RGB or
BI_BITFIELDS
DWORD &compression = bmInfo.bmiHeader.biCompression;
if( nHeight < 0 || (compression!=BI_RGB &&
compression!=BI_BITFIELDS))
return NULL;

LPVOID lpDIBBits;
if( bmInfo.bmiHeader.biBitCount > 8 )
lpDIBBits = (LPVOID)((LPDWORD)(bmInfo.bmiColors +
bmInfo.bmiHeader.biClrUsed) +
((compression == BI_BITFIELDS) ? 3 : 0));
else
lpDIBBits = (LPVOID)(bmInfo.bmiColors + nColors);


// Compute the cosine and sine only once
float cosine = (float)cos(radians);
float sine = (float)sin(radians);

// Compute dimensions of the resulting bitmap
// First get the coordinates of the 3 corners other than origin
int x1 = (int)(-nHeight * sine);
int y1 = (int)(nHeight * cosine);
int x2 = (int)(nWidth * cosine - nHeight * sine);
int y2 = (int)(nHeight * cosine + nWidth * sine);
int x3 = (int)(nWidth * cosine);
int y3 = (int)(nWidth * sine);

int minx = min(0,min(x1, min(x2,x3)));
int miny = min(0,min(y1, min(y2,y3)));
int maxx = max(x1, max(x2,x3));
int maxy = max(y1, max(y2,y3));

int w = maxx - minx;
int h = maxy - miny;


// Create a DIB to hold the result
int nResultRowBytes = ((((w * bpp) + 31) & ~31) / 8);
long len = nResultRowBytes * h;
int nHeaderSize = ((LPBYTE)lpDIBBits-(LPBYTE)hDIB) ;
HANDLE hDIBResult = GlobalAlloc(GMEM_FIXED,len+nHeaderSize);
// Initialize the header information
memcpy( (void*)hDIBResult, (void*)hDIB, nHeaderSize);
BITMAPINFO &bmInfoResult = *(LPBITMAPINFO)hDIBResult ;
bmInfoResult.bmiHeader.biWidth = w;
bmInfoResult.bmiHeader.biHeight = h;
bmInfoResult.bmiHeader.biSizeImage = len;

LPVOID lpDIBBitsResult = (LPVOID)((LPBYTE)hDIBResult + nHeaderSize);

// Get the back color value (index)
ZeroMemory( lpDIBBitsResult, len );
DWORD dwBackColor;
switch(bpp)
{
case 1: //Monochrome
if( clrBack == RGB(255,255,255) )
memset( lpDIBBitsResult, 0xff, len );
break;
case 4:
case 8: //Search the color table
int i;
for(i = 0; i < nColors; i++ )
{
if( bmInfo.bmiColors[i].rgbBlue == GetBValue(clrBack)
&& bmInfo.bmiColors[i].rgbGreen == GetGValue(clrBack)
&& bmInfo.bmiColors[i].rgbRed == GetRValue(clrBack) )
{
if(bpp==4) i = i | i<<4;
memset( lpDIBBitsResult, i, len );
break;
}
}
// If not match found the color remains black
break;
case 16:
// Windows95 supports 5 bits each for all colors or 5 bits for red
& blue
// and 6 bits for green - Check the color mask for RGB555 or RGB565
if( *((DWORD*)bmInfo.bmiColors) == 0x7c00 )
{
// Bitmap is RGB555
dwBackColor = ((GetRValue(clrBack)>>3) << 10) +
((GetRValue(clrBack)>>3) << 5) +
(GetBValue(clrBack)>>3) ;
}
else
{
// Bitmap is RGB565
dwBackColor = ((GetRValue(clrBack)>>3) << 11) +
((GetRValue(clrBack)>>2) << 5) +
(GetBValue(clrBack)>>3) ;
}
break;
case 24:
case 32:
dwBackColor = (((DWORD)GetRValue(clrBack)) << 16) |
(((DWORD)GetGValue(clrBack)) << 8) |
(((DWORD)GetBValue(clrBack)));
break;
}


// Now do the actual rotating - a pixel at a time
// Computing the destination point for each source point
// will leave a few pixels that do not get covered
// So we use a reverse transform - e.i. compute the source point
// for each destination point

for( int y = 0; y < h; y++ )
{
for( int x = 0; x < w; x++ )
{
int sourcex = (int)((x+minx)*cosine + (y+miny)*sine);
int sourcey = (int)((y+miny)*cosine - (x+minx)*sine);
if( sourcex >= 0 && sourcex < nWidth && sourcey
>= 0
&& sourcey < nHeight )
{
// Set the destination pixel
switch(bpp)
{
BYTE mask;
case 1: //Monochrome
mask = *((LPBYTE)lpDIBBits + nRowBytes*sourcey +
sourcex/8) & (0x80 >> sourcex%8);
//Adjust mask for destination bitmap
mask = mask ? (0x80 >> x%8) : 0;
*((LPBYTE)lpDIBBitsResult + nResultRowBytes*(y) +
(x/8)) &= ~(0x80 >> x%8);
*((LPBYTE)lpDIBBitsResult + nResultRowBytes*(y) +
(x/8)) |= mask;
break;
case 4:
mask = *((LPBYTE)lpDIBBits + nRowBytes*sourcey +
sourcex/2) & ((sourcex&1) ? 0x0f : 0xf0);
//Adjust mask for destination bitmap
if( (sourcex&1) != (x&1) )
mask = (mask&0xf0) ? (mask>>4) : (mask<<4);
*((LPBYTE)lpDIBBitsResult + nResultRowBytes*(y) +
(x/2)) &= ~((x&1) ? 0x0f : 0xf0);
*((LPBYTE)lpDIBBitsResult + nResultRowBytes*(y) +
(x/2)) |= mask;
break;
case 8:
BYTE pixel ;
pixel = *((LPBYTE)lpDIBBits + nRowBytes*sourcey +
sourcex);
*((LPBYTE)lpDIBBitsResult + nResultRowBytes*(y) +
(x)) = pixel;
break;
case 16:
DWORD dwPixel;
dwPixel = *((LPWORD)((LPBYTE)lpDIBBits +
nRowBytes*sourcey + sourcex*2));
*((LPWORD)((LPBYTE)lpDIBBitsResult +
nResultRowBytes*y + x*2)) = (WORD)dwPixel;
break;
case 24:
dwPixel = *((LPDWORD)((LPBYTE)lpDIBBits +
nRowBytes*sourcey + sourcex*3)) & 0xffffff;
*((LPDWORD)((LPBYTE)lpDIBBitsResult +
nResultRowBytes*y + x*3)) |= dwPixel;
break;
case 32:
dwPixel = *((LPDWORD)((LPBYTE)lpDIBBits +
nRowBytes*sourcey + sourcex*4));
*((LPDWORD)((LPBYTE)lpDIBBitsResult +
nResultRowBytes*y + x*4)) = dwPixel;
}
}
else
{
// Draw the background "color."
tppabs="http://www.codeguru.com/bitmap/color." The
background color
// has already been drawn for 8 bits per pixel and less
switch(bpp)
{
case 16:
*((LPWORD)((LPBYTE)lpDIBBitsResult +
nResultRowBytes*y + x*2)) =
(WORD)dwBackColor;
break;
case 24:
*((LPDWORD)((LPBYTE)lpDIBBitsResult +
nResultRowBytes*y + x*3)) |= dwBackColor;
break;
case 32:
*((LPDWORD)((LPBYTE)lpDIBBitsResult +
nResultRowBytes*y + x*4)) = dwBackColor;
break;
}
}
}
}

return hDIBResult;
}
iamwjp 2006-05-09
  • 打赏
  • 举报
回复
函数二: GetRotatedBitmap()使用 GetPixel & SetPixel

// GetRotatedBitmap - Create a new bitmap with rotated image
// Returns - Returns new bitmap with rotated image
// hBitmap - Bitmap to rotate
// radians - Angle of rotation in radians
// clrBack - Color of pixels in the resulting bitmap that do
// not get covered by source pixels
// Note - If the bitmap uses colors not in the system palette
// then the result is unexpected. You can fix this by
// adding an argument for the logical palette.
HBITMAP GetRotatedBitmap( HBITMAP hBitmap, float radians, COLORREF clrBack )
{
// Create a memory DC compatible with the display
CDC sourceDC, destDC;
sourceDC.CreateCompatibleDC( NULL );
destDC.CreateCompatibleDC( NULL );

// Get logical coordinates
BITMAP bm;
::GetObject( hBitmap, sizeof( bm ), &bm );

float cosine = (float)cos(radians);
float sine = (float)sin(radians);

// Compute dimensions of the resulting bitmap
// First get the coordinates of the 3 corners other than origin
int x1 = (int)(-bm.bmHeight * sine);
int y1 = (int)(bm.bmHeight * cosine);
int x2 = (int)(bm.bmWidth * cosine - bm.bmHeight * sine);
int y2 = (int)(bm.bmHeight * cosine + bm.bmWidth * sine);
int x3 = (int)(bm.bmWidth * cosine);
int y3 = (int)(bm.bmWidth * sine);

int minx = min(0,min(x1, min(x2,x3)));
int miny = min(0,min(y1, min(y2,y3)));
int maxx = max(x1, max(x2,x3));
int maxy = max(y1, max(y2,y3));

int w = maxx - minx;
int h = maxy - miny;


// Create a bitmap to hold the result
HBITMAP hbmResult = ::CreateCompatibleBitmap(CClientDC(NULL), w, h);

HBITMAP hbmOldSource = (HBITMAP)::SelectObject( sourceDC.m_hDC, hBitmap );
HBITMAP hbmOldDest = (HBITMAP)::SelectObject( destDC.m_hDC, hbmResult );

// Draw the background color before we change mapping mode
HBRUSH hbrBack = CreateSolidBrush( clrBack );
HBRUSH hbrOld = (HBRUSH)::SelectObject( destDC.m_hDC, hbrBack );
destDC.PatBlt( 0, 0, w, h, PATCOPY );
::DeleteObject( ::SelectObject( destDC.m_hDC, hbrOld ) );

// Set mapping mode so that +ve y axis is upwords
sourceDC.SetMapMode(MM_ISOTROPIC);
sourceDC.SetWindowExt(1,1);
sourceDC.SetViewportExt(1,-1);
sourceDC.SetViewportOrg(0, bm.bmHeight-1);

destDC.SetMapMode(MM_ISOTROPIC);
destDC.SetWindowExt(1,1);
destDC.SetViewportExt(1,-1);
destDC.SetWindowOrg(minx, maxy);

// Now do the actual rotating - a pixel at a time
// Computing the destination point for each source point
// will leave a few pixels that do not get covered
// So we use a reverse transform - e.i. compute the source point
// for each destination point

for( int y = miny; y < maxy; y++ )
{
for( int x = minx; x < maxx; x++ )
{
int sourcex = (int)(x*cosine + y*sine);
int sourcey = (int)(y*cosine - x*sine);
if( sourcex >= 0 && sourcex < bm.bmWidth && sourcey >= 0
&& sourcey < bm.bmHeight )
destDC.SetPixel(x,y,sourceDC.GetPixel(sourcex,sourcey));
}
}

// Restore DCs
::SelectObject( sourceDC.m_hDC, hbmOldSource );
::SelectObject( destDC.m_hDC, hbmOldDest );

return hbmResult;
}

iamwjp 2006-05-09
  • 打赏
  • 举报
回复
旋转图像

如果你的应用程序仅工作在Windows NT下,那么你可以通过API函数旋转你的位图。
你或者使用world transformation和BitBlt()或者使用PlgBlt()旋转位图。一个
使用第一种方法的函数显示在下面。

如果你的目标是多平台的,那么你的任务变得非常困难。你只能通过旋转源位图中
每个象素或者直接操作DIB字节得到旋转位图。第一种方法通过每个点的处理是非
常慢的,第二种方法是很复杂的,但它有足够快的速度。注:下面的所有函数旋转
后产生新的位图,如果你需要直接绘制位图,请自已修改函数。 其中函数1仅工作
在NT环境下,它是最简单也是最快的,可惜它不支持Windows95。

所有的函数所接受的角度单位是弧度,如果是角度单位是度请用下面的公式转换。

radian = (2*pi *degree)/360

旋转步骤:

创建一对与设备兼容的显示设备。一个用于源位图,一个用于旋转后的目标位图。

预计算正弦和余弦函数值,这样可以避免重复计算。

用下面的公式计算旋转图像后的矩形
newx = x.cos(angle) + y.sin(angle)
newy = y.cos(angle) - x.sin(angle)

旋转后的位图将不能占用整个新位图,我们将用背景色填充它。

点阵转换公式
newx = x * eM11 + y * eM21 + eDx
newy = x * eM12 + y * eM22 + eDy
其中eM11和eM22是角度的余弦值,eM21是角度的正弦,eM12是eM21的负值。 eDx & eDy
目的是旋转后的位图在新的位图不被剪切。

函数一:适用于NT

// GetRotatedBitmapNT - Create a new bitmap with rotated image
// Returns - Returns new bitmap with rotated image
// hBitmap - Bitmap to rotate
// radians - Angle of rotation in radians
// clrBack - Color of pixels in the resulting bitmap that do
// not get covered by source pixels
HBITMAP GetRotatedBitmapNT( HBITMAP hBitmap, float radians, COLORREF clrBack )
{
// Create a memory DC compatible with the display
CDC sourceDC, destDC;
sourceDC.CreateCompatibleDC( NULL );
destDC.CreateCompatibleDC( NULL );

// Get logical coordinates
BITMAP bm;
::GetObject( hBitmap, sizeof( bm ), &bm );

float cosine = (float)cos(radians);
float sine = (float)sin(radians);

// Compute dimensions of the resulting bitmap
// First get the coordinates of the 3 corners other than origin
int x1 = (int)(bm.bmHeight * sine);
int y1 = (int)(bm.bmHeight * cosine);
int x2 = (int)(bm.bmWidth * cosine + bm.bmHeight * sine);
int y2 = (int)(bm.bmHeight * cosine - bm.bmWidth * sine);
int x3 = (int)(bm.bmWidth * cosine);
int y3 = (int)(-bm.bmWidth * sine);

int minx = min(0,min(x1, min(x2,x3)));
int miny = min(0,min(y1, min(y2,y3)));
int maxx = max(0,max(x1, max(x2,x3)));
int maxy = max(0,max(y1, max(y2,y3)));

int w = maxx - minx;
int h = maxy - miny;

// Create a bitmap to hold the result
HBITMAP hbmResult = ::CreateCompatibleBitmap(CClientDC(NULL), w, h);

HBITMAP hbmOldSource = (HBITMAP)::SelectObject( sourceDC.m_hDC, hBitmap );
HBITMAP hbmOldDest = (HBITMAP)::SelectObject( destDC.m_hDC, hbmResult );

// Draw the background color before we change mapping mode
HBRUSH hbrBack = CreateSolidBrush( clrBack );
HBRUSH hbrOld = (HBRUSH)::SelectObject( destDC.m_hDC, hbrBack );
destDC.PatBlt( 0, 0, w, h, PATCOPY );
::DeleteObject( ::SelectObject( destDC.m_hDC, hbrOld ) );

// We will use world transform to rotate the bitmap
SetGraphicsMode(destDC.m_hDC, GM_ADVANCED);
XFORM xform;
xform.eM11 = cosine;
xform.eM12 = -sine;
xform.eM21 = sine;
xform.eM22 = cosine;
xform.eDx = (float)-minx;
xform.eDy = (float)-miny;

SetWorldTransform( destDC.m_hDC, &xform );

// Now do the actual rotating - a pixel at a time
destDC.BitBlt(0,0,bm.bmWidth, bm.bmHeight, &sourceDC, 0, 0, SRCCOPY );

// Restore DCs
::SelectObject( sourceDC.m_hDC, hbmOldSource );
::SelectObject( destDC.m_hDC, hbmOldDest );

return hbmResult;
}

smzh8 2006-05-09
  • 打赏
  • 举报
回复
用API的PlgBit
lixiaosan 2006-05-09
  • 打赏
  • 举报
回复
GDI+

http://www.yesky.com/282/1772782_1.shtml

19,468

社区成员

发帖
与我相关
我的任务
社区描述
VC/MFC 图形处理/算法
社区管理员
  • 图形处理/算法社区
加入社区
  • 近7日
  • 近30日
  • 至今
社区公告
暂无公告

试试用AI创作助手写篇文章吧