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 following?

sum  = 4 * 255 + A;
sum += 3 * 255 + R;
sum += 2 * 255 + G;
sum += 1 * 255 + B;

Is the value really unique?

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;

---

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.

Answer 2

You are trying to do a base convert or something of that sort. Anyway the logic is like in base converting. 4 bytes = 32 bit. So 32 bit unsigned integer would do well.

In this case, you have:

ARGB = A<<24 + R<<16 + G<<8 + B

it's like this:
you have 4 bytes of data, meaning

xxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxx

where X is either 1 or 0 valued bit. You map them like this:

AAAAAAAA RRRRRRRR GGGGGGGG BBBBBBBB

and then all you have to do is to add them, but before that you shift the bits. You shift the A bits to the left, by 8*3 (to be beyond the limits of R, G and B bits), then shift the R bits by 8*2, and so on.

You end up adding these 32 bit integers:

AAAAAAAA 00000000 00000000 00000000
00000000 RRRRRRRR 00000000 00000000
00000000 00000000 GGGGGGGG 00000000
00000000 00000000 00000000 BBBBBBBB

Where A, R, G, B can be either 0 or 1, and represent as a whole, the 8 bit value of the channel. Then you simply add them, and obtain the result. Or as DarkDust wrote, use not the + operator, but instead the | (bitwise or) operator, since it should be faster in this particular case.

Answer 3

#include<bitset>
void printBits(string s, int x)
{
    bitset <64> b(x);
    cout<<"\n"<<s<< " " << b;
}
long RGB(int a, int r, int g, int b)
{
    printBits("alpha    ", a);
    printBits("red      ", r);
    printBits("green    ", g);
    printBits("blue     ", b);

    long l = 0;

    l |= a << 24;
    l |= r << 16;
    l |= g << 8;
    l |= b;

    printBits("packed ARGB", l);

    return l;
}

Answer 4

You could do this:

Assuming a, r, g and b to be of type unsigned char/uint8_t:

uint32_t color = 0;
color |= a << 24;
color |= r << 16;
color |= g << 8;
color |= b;

Or more general (a, r, g and b being of any integer type):

uint32_t color = 0;
color |= (a & 255) << 24;
color |= (r & 255) << 16;
color |= (g & 255) << 8;
color |= (b & 255);

This will give you a unique integer for every ARGB combination. You can get the values back like this:

a = (color >> 24) & 255;
r = (color >> 16) & 255;
g = (color >> 8) & 255;
b = color & 255;