我想从鼠标光标获取位图,但是使用下一个代码,我只是无法获取颜色。鼠标光标位图
CURSORINFO cursorInfo = { 0 };
cursorInfo.cbSize = sizeof(cursorInfo);
if (GetCursorInfo(&cursorInfo)) {
ICONINFO ii = {0};
int p = GetIconInfo(cursorInfo.hCursor, &ii);
// get screen
HDC dc = GetDC(NULL);
HDC memDC = CreateCompatibleDC(dc);
//SelectObject(memDC, ii.hbmColor);
int counter = 0;
//
byte* bits[1000];// = new byte[w * 4];
BITMAPINFO bmi;
memset(&bmi, 0, sizeof(BITMAPINFO));
bmi.bmiHeader.biSize = sizeof(BITMAPINFOHEADER);
bmi.bmiHeader.biWidth = 16;
bmi.bmiHeader.biHeight = 16;
bmi.bmiHeader.biBitCount = 32;
bmi.bmiHeader.biPlanes = 1;
bmi.bmiHeader.biCompression = BI_RGB;
bmi.bmiHeader.biSizeImage = 0;
bmi.bmiHeader.biXPelsPerMeter = 0;
bmi.bmiHeader.biYPelsPerMeter = 0;
bmi.bmiHeader.biClrUsed = 0;
bmi.bmiHeader.biClrImportant = 0;
int rv = ::GetDIBits(memDC, ii.hbmColor, 0, 1, (void**)&bits, &bmi, DIB_RGB_COLORS);
}
在这种情况下,“BITMAPINFO”不会为全部调色板保留空间吗?为什么'GetDIBits'会写入它(尤其是'DIB_RGB_COLORS')? – jamesdlin 2012-06-07 22:45:42
@ jamesdlin,如果图像不需要调色板,那么你是对的,没关系。尽管安全,但这并不是一个坏主意。 – 2012-06-07 22:55:49