如何创建任意颜色深度的位图8 16 24 32位？

[yangege]

我的显示卡支持32位，一个24位的dib通过下面的函数。
转换为HBITMAP却变成了32位的位图。能否保持24位不变。
或者能转换成8 16 24位的位图（HBITMAP）？
HBITMAP CBitmapCtrl::DIBToDDB(HANDLE hDIB)
{
LPBITMAPINFOHEADER lpbi;
HBITMAP hbm;
CPalette pal;
CPalette* pOldPal;
CClientDC dc(NULL);

if (hDIB == NULL)
return NULL;

lpbi = (LPBITMAPINFOHEADER)hDIB;

int nColors = lpbi->biClrUsed ? lpbi->biClrUsed :
1 << lpbi->biBitCount;

BITMAPINFO &bmInfo = *(LPBITMAPINFO)hDIB ;
LPVOID lpDIBBits;
if( bmInfo.bmiHeader.biBitCount > 8 )
lpDIBBits = (LPVOID)((LPDWORD)(bmInfo.bmiColors +
bmInfo.bmiHeader.biClrUsed) +
((bmInfo.bmiHeader.biCompression == BI_BITFIELDS) ? 3 : 0));
else
lpDIBBits = (LPVOID)(bmInfo.bmiColors + nColors);

// Create and select a logical palette if needed
if( nColors <= 256 && dc.GetDeviceCaps(RASTERCAPS) & RC_PALETTE)
{
UINT nSize = sizeof(LOGPALETTE) + (sizeof(PALETTEENTRY) * nColors);
LOGPALETTE *pLP = (LOGPALETTE *) new BYTE[nSize];

pLP->palVersion = 0x300;
pLP->palNumEntries = nColors;

for( int i=0; i < nColors; i++)
{
pLP->palPalEntry[i].peRed = bmInfo.bmiColors[i].rgbRed;
pLP->palPalEntry[i].peGreen = bmInfo.bmiColors[i].rgbGreen;
pLP->palPalEntry[i].peBlue = bmInfo.bmiColors[i].rgbBlue;
pLP->palPalEntry[i].peFlags = 0;
}

pal.CreatePalette( pLP );

delete[] pLP;

// Select and realize the palette
pOldPal = dc.SelectPalette( &pal, FALSE );
dc.RealizePalette();
}


hbm = CreateDIBitmap(dc.GetSafeHdc(), // handle to device context
(LPBITMAPINFOHEADER)lpbi, // pointer to bitmap info header
(LONG)CBM_INIT, // initialization flag
lpDIBBits, // pointer to initialization data
(LPBITMAPINFO)lpbi, // pointer to bitmap info
DIB_RGB_COLORS ); // color-data usage

if (pal.GetSafeHandle())
dc.SelectPalette(pOldPal,FALSE);

return hbm;
}