RGB888 to RGB565 / Bit Shifting
bit-manipulation, c, rgb
Solution
rgb = ((r & 0b11111000) << 8) | ((g & 0b11111100) << 3) | (b >> 3);
We shift `r` left by 11-3=8 bits, `g` left by 5-2=3 bits and bitwise OR these with `b` shifted right by 3 bits.
Problem
I want to combine three characters into a short using bit shifting. This is for implementing the RGB565 color palette (where there are 5 bits for red, 6 for green, 5 for blue). Here is my example program, i'm just missing a step in the middle i think where i need to do some anding. ``` #include <stdio.h> int main( ){ unsigned char r, g, b; unsigned short rgb; r = 255; // 0xFF 1111 1111 g = 100; // 0x64 0110 0100 b = 50; // 0x32 0011 0010 r = r >> 3; // 0x31 0001 1111 g = g >> 2; // 0x19 0001 1001 b = b >> 3; // 0x06 0000 0110 //r = r & something; // //g = g & something; // //b = b & something; // // Desired result: // R G B // 0xFB26 11111 011001 00110 rgb = r | g | b; printf( "r 0x%x g 0x%x b 0x%x, rgb 0x%08x\n", r, g, b, rgb ); } ``` You can see my desired result at the end. Thanks for the help!