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bitmaskin 16bit color value

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I though i knew how to use bitmask until i tryed this code:
#define RGB16(red, green, blue) ( ((red >> 3) << 11) | ((green >> 2) << 5) | (blue >> 3))  

USHORT col = RGB16(255,255,255);

USHORT r,g,b;

  r = ((col & 0x1F)<<3);
  g = ((col>>5 & 0x3F)<<2);
  b = (((col>>11) & 0x1F)<<3);
The results of r g and be are bad they arent 255 like they should, what do i do wrong? [edited by - Craazer on October 22, 2002 6:09:19 PM]

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Original post by Prototype
Shouldn''t the second line be

USHORT col = RGB16(255,255,255); ?

That will use the macro you defined.

Oh that was just typo, and it wasnt the real problem.

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I can see a couple of possible problems there. First, you seem to have red and blue the wrong way round in the last three lines. The macro does ((red >> 3) << 11), so the corresponding "unpack" expression would be (((col>>11) & 0x1F)<<3) rather than ((col & 0x1F)<<3).

The second problem is that you are packing three bytes into a word - 24 bits into 16. The macro drops the 3 least significant bits of red and blue and the two least significant bits of green. That being the case, I''d expect you got:

r = 0b11111000 = 0xF8 = 248
g = 0b11111100 = 0xFC = 252
b = 0b11111000 = 0xF8 = 248

In other words, the missing bits are just set to zero. I''m not sure if this is the correct way to fix this, but I seem to remember seeing the upper bits of a component used to fill the missing ones, something like this:

r = ((col & 0xF800) >> 8) | ((col & 0xF800) >> 13);
g = ((col & 0x07E0) >> 5) | ((col & 0x07E0) >> 8);
b = ((col & 0x001F) << 3) | ((col & 0x001F) >> 2);

It doesn''t get back the lost bits, but it means that RGB16(255, 255, 255) gives pure white if you unpack the components.

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