How to get 8 bit Colors?

Question

So ,at this moment, I got a method which generates 3 random integers, from 0 to 255.

I use the 3 integers for colors.Red, Green ,Blue.

So,at this moment, if I want to set the color of something, with the generated colors, I use this:

 Color.argb(255, r, g, b); //255 =max intensity and after ,there are red green blue

What I need to do ,is to convert the 3 integers ,or eventually the intensity value too, to a 8bit integer.

Any sort of documentation or guidance is highly appreciated!

If more information is needed, I will comment or modify the body of the question.

hava a look here, http://stackoverflow.com/questions/14775442/relative-8-bit-color-of-a-16-bit-conversion — user902383, Dec 04 '14 at 09:58
To put 24 bits of data in 8 bits of space you will loose information. Some systems use a palette (a lookup table of colours) to get as close as they can to the original (gifs can do this). — Andy Newman, Dec 04 '14 at 10:08
see http://stackoverflow.com/questions/4801366/convert-rgb-values-into-integer-pixel — sravtej, Dec 04 '14 at 10:08
Using 8-bit will lose some data about the color anyway. That's the "beauty" of the 64 colors which you can get. — , Dec 04 '14 at 10:15

score 8 · Accepted Answer · 2014-12-04T10:39:31.297

You can encode the colors the following way:

Bits    0### 1### 2### 3### 4### 5### 6### 7###
        Alpha---- Red------ Green---- Blue-----

Note that you'll lose a whole lot of information about the color (but I think that's the thing you want to get).

Things you'll need to do in order to encode:

Change the range of colors (0-255 to 0-3)
Shift the colors properly and add them to get the 8bit value.

Here's some example code:

import java.awt.Color;

abstract class EightBit {
  public static int fromColor(Color c) {
    return ((c.getAlpha() >> 6) << 6)
         + ((c.getRed()   >> 6) << 4)
         + ((c.getGreen() >> 6) << 2)
         +  (c.getBlue()  >> 6);
  }
  public static Color toColor(int i) {
    return new Color(((i >> 4) % 4) * 64,
                     ((i >> 2) % 4) * 64,
                      (i       % 4) * 64,
                      (i >> 6)      * 64);
  }
}

Explanation

Encoding

Let's start with an example color: new Color(200, 59, 148, 72). Now we'll convert that into an integer. The binary representation of the color is:

Alpha 200 -- 11001000
Red    59 -- 00111011
Green 148 -- 10010100
Blue   72 -- 01001000

Now, we shift them to the right by 6 bits (so we get the first 2 bits):

Alpha 3 -- 11
Red   0 -- 00
Green 2 -- 10
Blue  1 -- 01

Now we put them together:

Bits  [ 1 ][ 1 ][ 0 ][ 0 ][ 1 ][ 0 ][ 0 ][ 1 ] -- 209
       ALPHA---  RED-----  GREEN---  BLUE----

It's 209. See?

Decoding

So we're back at our 8bit number: 209. We want to decode it. First, we need to get the 2-bit colors back by shifting them to the right, and modulo 4:

Bits  [ 1 ][ 1 ][ 0 ][ 0 ][ 1 ][ 0 ][ 0 ][ 1 ]
      \_shift_by_6_bits_____________[ 1 ][ 1 ] -- 3 (Alpha)
                \_by_4_bits_________[ 0 ][ 0 ] -- 0 (Red)
                          \_by_2____[ 1 ][ 0 ] -- 2 (Green)
                   shift by 0 bits: [ 0 ][ 1 ] -- 1 (Blue)

Now we multiply them by 64:

3 * 64 = 192 (Alpha)
0 * 64 =   0 (Red)
2 * 64 = 128 (Green)
1 * 64 =  64 (Blue)

And put them back into a Color object. As you can see, the colors are different: some information about the color was lost in the process. This is called lossy compression.

Would you please explain me the whole process? I am pretty sure I understand how it works, but I want to make sure I don't use misunderstood concepts. Anyways, overall it looks like what I need. Going to test it right away and check the answer if everything is right. (Gonna give you an up too, as long as you explain what you have done,a bit) — Vlad, Dec 04 '14 at 10:19
Explanation is done! Make sure to accept the answer (with its exhaustive explanation). — , Dec 04 '14 at 10:40

How to get 8 bit Colors?

1 Answers1

Explanation

Encoding

Decoding