为什么MS Example of GetDIBits会创建一个新的BITMAPINFOHEADER？

Question

使用GetDIBits here的微软示例必须编写以下代码。我的问题是，为什么他们要创建一个新的bi项，而不是只在调用GetDIBits时使用bmpScreen ?：

    // Get the BITMAP from the HBITMAP
    GetObject(hbmScreen,sizeof(BITMAP),&bmpScreen);
     
    BITMAPFILEHEADER   bmfHeader;    
    BITMAPINFOHEADER   bi;
     
    bi.biSize = sizeof(BITMAPINFOHEADER);    
    bi.biWidth = bmpScreen.bmWidth;    
    bi.biHeight = bmpScreen.bmHeight;  
    bi.biPlanes = 1;    
    bi.biBitCount = 32;    
    bi.biCompression = BI_RGB;    
    bi.biSizeImage = 0;  
    bi.biXPelsPerMeter = 0;    
    bi.biYPelsPerMeter = 0;    
    bi.biClrUsed = 0;    
    bi.biClrImportant = 0;

    DWORD dwBmpSize = ((bmpScreen.bmWidth * bi.biBitCount + 31) / 32) * 4 * bmpScreen.bmHeight;

    // Starting with 32-bit Windows, GlobalAlloc and LocalAlloc are implemented as wrapper functions that 
    // call HeapAlloc using a handle to the process's default heap. Therefore, GlobalAlloc and LocalAlloc 
    // have greater overhead than HeapAlloc.
    HANDLE hDIB = GlobalAlloc(GHND,dwBmpSize); 
    char *lpbitmap = (char *)GlobalLock(hDIB);    

    // Gets the "bits" from the bitmap and copies them into a buffer 
    // which is pointed to by lpbitmap.
    GetDIBits(hdcWindow, hbmScreen, 0,
        (UINT)bmpScreen.bmHeight,
        lpbitmap,
        (BITMAPINFO *)&bi, DIB_RGB_COLORS);

Answer 1

GetDIBits

如果lpvBits参数是有效指针，则必须初始化BITMAPINFOHEADER结构的前六个成员以指定DIB的大小和格式。扫描线必须在DWORD上对齐，但RLE压缩的位图除外。