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# 16-bit Color

By TANSTAAFL | Published Nov 23 1999 09:29 AM in DirectX and XNA

dword color bits ddpixelformat format 1)==0 looked dwbbitmask ddpf
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Today, I'm going to address a problem that you might be dealing with, or dealing with in the near future… rectifying a 24-bit image onto a 16-bit DirectDraw surface.

A few days ago, I was working on my old computer, an ATI VGA Wonder. It had 5 bits for each R,G, or B value(using only 15 bits), so the pixel format looked something like this

XRRRRRGGGGGBBBBB

Now, my new computer has a better video card. It uses all 16 bits, making the pixel format look like:

RRRRRGGGGGGBBBBB

So, I got to thinking… how am I going to make it so that I can specify any color to put on the screen, and have it turn out right, no matter what video card it is.

As a side note, I have written my own bitmap loader to read in 24 bit BMPs, and rectify them to a 16 bit surface (I'll probably share it at some later time, most likely in a separate article)

I sat down and set about my task. The first thing I did was open up the DirectX.hlp file, and took a look at the IDirectDrawSurface Interface. One of its member functions was called "GetPixelFormat". Okay. I had something to work with.

GetPixelFormat takes a pointer to a DDPIXELFORMAT structure (not surprisingly). I looked at the DDPIXELFORMAT structure, and I found a couple of fields that interested me… namely dwRBitMask, dwGBitMask, dwBBitMask.

Now I had all of the information I wanted, just not in a format I could use.

DDPIXELFORMAT ddpf;
memset(&ddpf,0,sizeof(DDPIXELFORMAT);
ddpf.dwSize=sizeof(DDPIXELFORMAT);
lpDDS->GetPixelFormat(&ddpf);

What now?

Well, in order to figure out how far to shift bits to the left, I had to do this:

//warning, this code is UGLY

DWORD dwBSHL=0;
DWORD dwRSHL=0;
DWORD dwGSHL=0;

{
dwBSHL++;
}

{
dwGSHL++;
}

{
dwRSHL++;
}

At this point, if we assumed that I had the correct number of bits stored in three UCHARs named Red, Green, and Blue, I could calculate the 16-bit color by doing this:

Color=Blue << dwBSHL + Green << dwGSHL + Red << dwRSHL;

However, right now, we have 8 bits in each of Red, Green, and Blue, and we need only 5 or 6. We have to be able to shift the bits of the colors to the right before shifting them to the left to make a 16-bit color.

So, let's determine how many bits to the right we have to shift them:

DWORD dwBSHR=8;
DWORD dwRSHR=8;
DWORD dwGSHR=8;

{
dwBSHR--;
}

{
dwGSHR--;
}

{
dwRSHR--;
}

Now, we can accurately calculate a 16bit color from a 24Bit color.

Color=(Blue >> dwBSHR) << dwBSHL + (Green >> dwGSHR) << dwGSHL + (Red >> dwRSHR) << dwRSHL;