16-bit Color

Another article covering 16-bit color? Considering how often the issue comes up, why not? Besides, it's by TANSTAAFL.

November 23, 1999 8.5k views

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 dwBMask=ddpf.dwBBitMask;
 DWORD dwRMask=ddpf.dwRBitMask;
 DWORD dwGMask=ddpf.dwGBitMask;
 
 DWORD dwBSHL=0;
 DWORD dwRSHL=0;
 DWORD dwGSHL=0;
 
 while((dwBMask & 1)==0)
 {
 dwBMask=dwBMask >> 1;
 dwBSHL++;
 }
 
 while((dwGMask & 1)==0)
 {
 dwGMask=dwGMask >> 1;
 dwGSHL++;
 }
 
 while((dwRMask & 1)==0)
 {
 dwRMask=dwRMask >> 1;
 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;
 
 while((dwBMask & 1)==1)
 {
 dwBMask=dwBMask >> 1;
 dwBSHR--;
 }
 
 while((dwGMask & 1)==1)
 {
 dwGMask=dwGMask >> 1;
 dwGSHR--;
 }
 
 while((dwRMask & 1)==1)
 {
 dwRMask=dwRMask >> 1;
 dwRSHR--;
 }