Started by Feb 12 2002 08:10 AM

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6 replies to this topic

Posted 12 February 2002 - 08:10 AM

howdy.. I''m trying to figure out how this macro works, I was wondering if anybody familiar with LaMothe''s books, has come across this before.
// builds up a 16bit color value
#define _RGB16BIT(r,g,b) ((b%32) +((g%64) << 5) + ((r%32) << 11))
i know what it''s doing.. but i''d prefer to know how it''s doing it, as i''m trying to improve my programming abilities. Specifically, I''d like to know what the mod32 does, as well as the bit-shifting. (hmm.. to clarify that, i know what the operators do, but i''d like to know what they''re doing in relation to building up a 16bit color)
If anyone has a comment, or could point me toward a source where i could learn more about how this works, it would be greatly appreciated
Ern

Posted 12 February 2002 - 08:45 AM

The modulo/modulus operator (`operator%`) returns the remainder of the left-hand side argument after division by the right-hand side. Thus, `1025 % 2` and `3 % 2` both return `1` - an elementary way to check if a number is odd, btw (using binary AND is better).

The bitwise shift operators move all bits in the value over the righ-hand argument number of times and fill in the "vacated" bits with zeroes. So`01011011 >> 4 => 00000101` and `01011011 << 2 => 01101100`. This bitshifting is a quick way to do multiplication by powers of two (`x * 2`^{n} = x << n).

So Andre''s macro is simply putting the low 5 bits of the`b` parameter (blue) as the low 5 bits of the resultant `RGB16BIT` value, the low 6 bits of `g` as the middle 6 green bits (by shifting it 5 places "up"), and the low 5 bits of `r` as the high 5 red bits (by shifting it 5 + 6 bits "up"). Why 5:6:5? 16-bit color mode. The human eye is slightly more sensitive to green, so video engineers decided to give an extra bit of definition to green in 16-bit mode.

*I wanna work for Microsoft!*

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Thanks to Kylotan for the idea!

The bitwise shift operators move all bits in the value over the righ-hand argument number of times and fill in the "vacated" bits with zeroes. So

So Andre''s macro is simply putting the low 5 bits of the

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Thanks to Kylotan for the idea!

Posted 12 February 2002 - 09:15 AM

thanks for the quick reply

just to clarify one final point..

modulo 32 (%32) is used to obtain the low 5 bits of say the b parameter.. why use 32 specifically to obtain the low 5 bits of b? i''m guessing because 32 = 2^5, and you use %64 to obtain the low 6 bits of green because 64 = 2^6. (sorry, about the font, i''m not familiar with the codes to get superscript etc)

Thanks again

Ern

just to clarify one final point..

modulo 32 (%32) is used to obtain the low 5 bits of say the b parameter.. why use 32 specifically to obtain the low 5 bits of b? i''m guessing because 32 = 2^5, and you use %64 to obtain the low 6 bits of green because 64 = 2^6. (sorry, about the font, i''m not familiar with the codes to get superscript etc)

Thanks again

Ern

Posted 12 February 2002 - 09:27 AM

quote:Original post by Baine

why use 32 specifically to obtain the low 5 bits of b? i''m guessing because 32 = 2^5, and you use %64 to obtain the low 6 bits of green because 64 = 2^6.

Exactly.

I use plain HTML to get super/subscript: <sup></sup>, <sub></sub>

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Thanks to Kylotan for the idea!

Posted 12 February 2002 - 09:44 AM

quote:Original post by Baine

why use 32 specifically to obtain the low 5 bits of b? i''m guessing because 32 = 2^5, and you use %64 to obtain the low 6 bits of green because 64 = 2^6.

You are correct.

By returning the remainer after a division, b%32 basically discards the upper bits of b, returning the lower 5 bits.

If any of the inputs to this macro are out of range, this code will silently truncate them to the correct range.

Personally, I would replace the modulus operator with the bitwise AND operator. It is more obvious to me that the upper bits are being masked out.

Shawn

Posted 12 February 2002 - 12:17 PM

umm, not to be mean but that method is utter bull dung. ist amazing that he got paid to write that. you NEVER use modulo for this operation. you want the upper bits, not the lower ones.

#define RGB565(r,g,b) ((r&0xF8)<<11)|((g&0xFC)<<5)|((b&0xF8)>>3))

#define RGB565(r,g,b) ((r&0xF8)<<11)|((g&0xFC)<<5)|((b&0xF8)>>3))