Converting a 16/24 bit hex value to R, G, B.
Author
I have seen in the tutorials many ways of converting R, G, and B values from 0-255 (24 bit) into a correct color value, but I was wondering how one would reverse the process?
Weel if the format for your card in 24 bit is R(8bits)G(8bits)B(8bits) (and it allmost certainly is) then if you have R 8bits, B 8bits and G 8 bits, you have to do something like:
unsigned char R,G,B;
_int32 Color = ((dword)R << 16) + ((dword)G << 8) + B
or
_int32 Color = ((dword)R << 16) & ((dword)G << 8) & (dword)B
You know, I never wanted to be a programmer...
Alexandre Moura
unsigned char R,G,B;
_int32 Color = ((dword)R << 16) + ((dword)G << 8) + B
or
_int32 Color = ((dword)R << 16) & ((dword)G << 8) & (dword)B
You know, I never wanted to be a programmer...
Alexandre Moura
unsigned int RGBValue,x;
unsigned char R,G,B;
x = RGBValue;
B = RGBValue & 0xff;
x >>=8;
G = x & 0xff;
x >>=8;
R = x & 0xff;
In the last line you can remove " & 0xff" if you''re sure that the highest byte of RGBValue is 0.
quote:Original post by alexmoura
Weel if the format for your card in 24 bit is R(8bits)G(8bits)B(8bits) (and it allmost certainly is) then if you have R 8bits, B 8bits and G 8 bits, you have to do something like:
unsigned char R,G,B;
_int32 Color = ((dword)R << 16) + ((dword)G << 8) + B
or
_int32 Color = ((dword)R << 16) & ((dword)G << 8) & (dword)B
I think Jabober wanted the inverse operation....
and you should write | instead of &.
You''re right I mistook both the & and the question 
(or should be 
?)
You know, I never wanted to be a programmer...
Alexandre Moura
(or should be ?) You know, I never wanted to be a programmer...
Alexandre Moura
inline DWORD RGB15to24( DWORD c ){ c = ((c & 0x1F)<<3) | ((c & 0x3E0)<<6) | ((c & 0x7C00)<<9); return c + ((c>>5) & 0x070707);}inline DWORD RGB16to24( DWORD c ){ c = ((c & 0x1F)<<3) | ((c & 0xF800)<<8); DWORD g = (c & 0x7E0)<<5); return c + ((c>>5) & 0x070007) + g + ((g>>6) & 0x000300);} This topic is closed to new replies.
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