Bit Shift and Bitwise operations to encode RGB values

Question

I would like to encode the RGB color to a single integer value.

Let's say the algorithm for encoding is like this:

int code = (blue * 256 * 256) + (green * 256) + red

How can encode/decode RGB components to/from the code using bit shift and/or bitwise operators?

The second answer in http://stackoverflow.com/questions/2615522/java-bufferedimage-getting-red-green-and-blue-individually just reverse the operations — arynaq, Oct 09 '13 at 16:04
Yes, but my code is 24 bit, why do I need to multiply each component by 0x000000FF? — Jan Krakora, Oct 09 '13 at 16:12
Because of the alpha, integer is 32 bit anyway so you can just set the alphaportion to zero. — arynaq, Oct 09 '13 at 16:13

Dev Blanked · Accepted Answer · 2013-10-09T16:47:00.203

12

    int blueMask = 0xFF0000, greenMask = 0xFF00, redMask = 0xFF;
    int r = 12, g = 13, b = 14;
    int bgrValue = (b << 16) + (g << 8) + r;
    System.out.println("blue:" + ((bgrValue & blueMask) >> 16));
    System.out.println("red:" + ((bgrValue & redMask)));
    System.out.println("green:" + ((bgrValue & greenMask) >> 8));

edited Oct 09 '13 at 16:47

answered Oct 09 '13 at 16:20

Dev Blanked

8,555
3
26
32

And how to encode individual RGB components to the integer value using bit shift? – Jan Krakora Oct 09 '13 at 16:24
@Behnil use (r << 16) + (g << 8) + b – Dev Blanked Oct 09 '13 at 16:27
actually the masks are int blueMask = 0xFF0000, greenMask = 0xFF00, redMask = 0xFF and shift by 16 is on the blue component instead of the red one – Jan Krakora Oct 09 '13 at 16:34
2

@Behnil yes u r correct i used it as RGB, but I see u r original question has it as BGR, but may be it's better to have it as RGB which is what most people refer to – Dev Blanked Oct 09 '13 at 16:37
Yes, if I could use RGB, I would use java.awt.Color. But I have to use BGR because that's how the data are stored in the RDBMS I connect to. – Jan Krakora Oct 09 '13 at 16:42
@Behnil have updated the answer to be inline with u r exact question – Dev Blanked Oct 09 '13 at 16:47

score 1 · Answer 2 · answered Oct 09 '13 at 16:29

If you simply want to/from RGB conversion and don't care how I would suggest using java.awt.Color

int r = 255; //red
int g = 255; //green
int b = 255; //blue
int a = 255; //alpha
Color c = new Color(r,g,b,a);

The using the getRGB method and getRed, getBlue, getGreen methods

int RGB = c.getRGB();
int red = c.getRed();
int blue = c.getBlue();
int green = c.getGreen();

Alternatively you can construct a color object using the Color(r,g,b) constructor, it will have default 255 alpha.

With bit operations (ARGB, 32 bit colorspace). Constructing the RGB color:

int alpha = 255;    
int red = 128;
int green = 128;
int blue = 128;
int RGB = (alpha << 24);
RGB = RGB | (red << 16);
RGB = RGB | (green << 8);
RGB = RGB | (blue);

System.out.println(Integer.toBinaryString(RGB));

Out 11111111100000001000000010000000

Decoding is done as in the link in the comment.

score 1 · Answer 3 · answered Oct 09 '13 at 16:38

Here is a mock program that I've done up that might assist you. I approached the conversion much like Dev Blanked based off an old program I did, but he answered while I was putting the program together. Since I did the work anyways, figured I'd share in case it helped in any way.

import java.util.Scanner;
import java.math.*;

public class RGB{

public static void main(String[]args){
    Scanner scan = new Scanner(System.in);
    int code; //Code for the color
    int red, green, blue; //Individual colors
    int rMask = 0xFF0000, gMask = 0xFF00, bMask = 0xFF; //Masks for the colors

    //Take input
    System.out.println("Please enter the red color. Range [0, 255] only please.");
    red = scan.nextInt();
    System.out.println("Please enter the green color. Range [0, 255] only please.");
    green = scan.nextInt();
    System.out.println("Please enter the blue color. Range [0, 255] only please.");
    blue = scan.nextInt();

    //Generate code based on Behnil's way.
    code = 0;
    code += (int) (red * Math.pow(2, 16));
    code += (int) (green * Math.pow(2, 8));
    code += (int) (blue * Math.pow(2,0));
    System.out.println("The code is " + code + ".");

    //Clear values
    red = 0;
    green = 0;
    blue = 0;

    //Obtain values.
    red = (code & rMask) >> 16;
    green = (code & gMask) >> 8;
    blue = (code & bMask);

    System.out.println("Your red value is: " + red);
    System.out.println("Your green value is: " + green);
    System.out.println("Your blue value is: " + blue);      
}

}

score 1 · Answer 4 · answered Mar 18 '15 at 18:55

public static void main(String[] args){
    int red = 111;
    int green = 222;
    int blue = 121;

    int code = red*256*256 + green*256 + blue;

    blue = code%256;
    green = (code%(256*256) - blue)/256;
    red = (code - blue - green*256)/(256*256); 

    System.out.println("" + red + green + blue);
}

this outputs 111222121 as intended. This is the way i fixed it but i am not sure if the pro's agree with this as it might be slower than using bitshifts

Bit Shift and Bitwise operations to encode RGB values

4 Answers4

Linked