Question

谁会用C++编程序画出图形，要求图行可以实现旋转，缩放，平移

Answer 1

缩放用：StretchBlt图片任意角度翻转：HBITMAP CBarDemoView::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/180 * PI);
float sine = (float)sin(radians/180 * PI); 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 ); nWidth[m_count] = bm.bmWidth;
nHeight[m_count] = bm.bmHeight; // Restore DCs
::SelectObject( sourceDC.m_hDC, hbmOldSource );
::SelectObject( destDC.m_hDC, hbmOldDest ); return hbmResult;
}