Is it possible to BitBlt directly from a GDI+ bitmap?

问题

Is it possible to use BitBlt to copy directly out of a GDI+ bitmap without using GetHBitmap?

GetHBitmap is slow because it makes a new copy of the whole image, in addition to and slower than the BitBlt copy, and the given HBITMAP must be disposed. The image is large.

Is there a way to point BitBlt to use the pixel data of the original GDI+ image?

EDIT: I can get a pointer to where the GDI+ bitmap pixel data is in the memory. Can I create an HBITMAP that points to the GDI+ bitmap pixel data to avoid the extra copy, and BitBlt from that?

回答1:

After searching for days, it suddenly hit me that the answer had been staring me in the face all the time! I was creating a GDI+ bitmap from a pointer to a byte array. Then trying to create an HBITMAP using the same pointer. But I could just as easily create the HBITMAP first and use the pointer from that to create the GDI+ bitmap.

It works like a charm! You can mix GDI and GDI+ operations however you like. The image is both plain GDI and GDI+ at once. Instead of using DrawImage, you can BitBlt from the exact same pixel data!

Here's the code:

// Create the HBITMAP
BITMAPINFO binfo = new BITMAPINFO();
binfo.biSize = (uint)Marshal.SizeOf(typeof(BITMAPINFO));
binfo.biWidth = width;
binfo.biHeight = height;
binfo.biBitCount = (ushort)Image.GetPixelFormatSize(pixelFormat);
binfo.biPlanes = 1;
binfo.biCompression = 0;

hDC = CreateCompatibleDC(IntPtr.Zero);

IntPtr pointer;
hBitmap = CreateDIBSection(hDC, ref binfo, 0, out pointer, IntPtr.Zero, 0);

// Create the GDI+ bitmap using the pointer returned from CreateDIBSection
gdiBitmap = new Bitmap(width, height, width * binfo.biBitCount >> 3, pixelFormat, pointer);

来源：https://stackoverflow.com/questions/4588132/is-it-possible-to-bitblt-directly-from-a-gdi-bitmap