How to invert an RGB color in integer form?

Question

Given an RGB color in 32-bit unsigned integer form (eg. 0xFF00FF), how would you invert it (get a negative color), without extracting its individual components using bitshift operations?

I wonder whether it's possible using just bitwise operations (AND, OR, XOR).

More precisely, what's the algorithm that uses the least number of instructions?

Answer 1

I think it is so simple. You can just calculate 0xFFFFFF-YourColor. It will be the inverted color.

int neg = 0xFFFFFF - originalRGB

// optional: set alpha to 255:
int neg = (0xFFFFFF - originalRGB) | 0xFF000000;

Answer 2

Use this method to invert each color and maintain original alpha.

int invert(int color) {
  return color ^ 0x00ffffff;
}

xor (^) with 0 returns the original value unmodified. xor with 0xff flips the bits. so in the above case we have 0xaarrggbb we are flipping/inverting r, g and b.

This should be the most efficient way to invert a color. arithmetic is (marginally) slower than this utterly simple bit-wise manipulation.

if you want to ignore original alpha, and just make it opaque, you can overwrite the alpha:

int invert(int color) {
  0xff000000 | ~color;
}

in this case we just flip every bit of color to inverse every channel including alpha, and then overwrite the alpha channel to opaque by forcing the first 8 bits high with 0xff000000.