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nodoubt

Macro question

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Hi, I am a little confused about the macros used to define a RGB color.. I''ve seen #define _RGB16BIT555(r,g,b) ((b & 31) + ((g&31) << 5) + ((r&31) << 10)) and #define _RGB16BIT555(r,g,b) ((r << 10) | (g << 5) | (b)) most of the text/docs just list them and assumed the reader knows why. Well, I''m sorry but I dont understand why you have to shift (divide) the values, and why by the amount (10, 5, etc). Can someone kind enough to explain?

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well, the << operator is a binary shift.

To my knowlege, it shifts the byte however many places to the left. ( >> does right). this means that:

0111 1111 >> 1;

is equal to 1111 1110. (see what happened? the last bit came from nowhere and we dropped the higest bit)

1111 1110 << 1 would be: 0111 1111

Note that this is all binary.

this is just my limited knowlege, though.

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Ehm, what are you talking about? You're right that >> shifts to the right, but then you write that 0111 1111 >> 1 is 1111 1110. Clearly, you have moved the bits to the left in that example, not to the right.

What the macro does is just to pack the bits into a 16 bit value, which has the format 0rrr rrgg gggb bbbb.

First the red color is shifted to the left ten steps:
r: 0000 0000 000r rrrr
r << 10: 0rrr rr00 0000 0000

And the green color five steps:
g: 0000 0000 000g gggg
g << 5: 0000 00gg ggg0 0000

Finally, put them together with or-statements '|':
0rrr rr00 0000 0000 or
0000 00gg ggg0 0000 or
0000 0000 000b bbbb
---------------------
0rrr rrgg gggb bbbb

And you have the packed rgb-value.

[edited by - aleph on December 5, 2002 2:46:10 PM]

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