How to pack ARGB to one integer uniquely?

Question

I have four integer values (0 - 255) for an ARGB color map.

Now I want to make a unique float or integer of these four integers. Is it possible to do it like the fol

Answer 1

Not quite. You need to use bit-shifting and not simple multiplication.

Each value in your color map is 8 bytes long, correct? So in order for the resulting number to be unique, it must string them all together, for a total of 8*4=32 bits. Look at the following:

You want to take:

AAAAAAAA
RRRRRRRR
GGGGGGGG
BBBBBBBB

and make it look like:

AAAAAAAARRRRRRRRGGGGGGGGBBBBBBBB

This means you have to add the following together:

AAAAAAAA000000000000000000000000
        RRRRRRRR0000000000000000
                GGGGGGGG00000000
                        BBBBBBBB
--------------------------------
AAAAAAAARRRRRRRRGGGGGGGGBBBBBBBB

We accomplish this by bit-shifting to the left. Taking A and shifting 24 bits to the left will produce AAAAAAAA followed by 24 0 bits, just like we want. Following that logic, you will want to do:

sum = A << 24 + R << 16 + G << 8 + B;

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To illustrate why what you suggest (using multiplication) does not work, what you suggest results in the following binary numbers, which you can see overlap:

255 * 1 = 0011111111
255 * 2 = 0111111110
255 * 3 = 1011111101
255 * 4 = 1111111100

Furthermore, simply adding your A, R, G, B values to the resulting number will always be constant. Simplifying your math above we get:

4 * 255 + A + 3 * 255 + R + 2 * 255 + G + 1 * 255 + B
255 * (4 + 3 + 2 + 1) + A + R + G + B
255 * (10) + A + R + G + B
2550 + A + R + G + B

Oops.