Not fully understanding this part of code

Question

I recently came accross this code and I can't get my head around it. Could someone, please, explain me what is happening there?

union Color32
{
   struct ARGB
   {
      uint8_t b;
      uint8_t g;
      uint8_t r;
      uint8_t a;
   } parts;
   uint32_t argb;
   Color32() : argb(0) {}
   Color32(uint32_t c_argb) : argb(c_argb) {}
   Color32(uint8_t a, uint8_t r, uint8_t g, uint8_t b)
   {
       parts.a=a;
       parts.r=r;
       parts.g=g;
       parts.b=b;
   }
};

From the constructors in the `union`, I think this might be C++, not C. — wolfPack88, Mar 19 '15 at 16:06
Reading material: http://stackoverflow.com/questions/346536/difference-between-a-structure-and-a-union-in-c, http://en.wikipedia.org/wiki/Union_type#C.2FC.2B.2B. — R Sahu, Mar 19 '15 at 16:07

Persixty · Accepted Answer · 2015-03-19T17:55:01.740

The members of a union share storage. This is in contrast to a struct where each member is allocated distinct storage.

The effect of this code is that the members b,g,r & a of the ARGB struct share storage with the other member argb of the Color32 union.

So the constructor that sets the value of argb implicitly sets the values of b,g,r & a as the ordered parts of the bit pattern of argb.

Conversely the b,g,r & a constructor builds up the value of argb.

The diagram under 'ARGB' here is a really good picture of how those parts are 'packed' into a 32 bit block.

http://en.wikipedia.org/wiki/RGBA_color_space

I probably don't need to explain that b stands for blue, ggreen , r for red and a for alpha (AKA Transparency)!

You may encounter difficulties if there is a requirement for big-endian /little-endian portability.

score 0 · Answer 2 · answered Mar 19 '15 at 16:09

ARGB is a structure of 4 uints which represents green, blue, red and for a I don't know. The structure is created with the name parts so when you have to use it you must act like: parts.r = 20 or parts.g = 50.

Then it create an object color which can contain the 4 parameters of the argb (4*8 = 32).

And it initialized the object color 32 with the 4 parameters that you give him : a,b,c,d

Color32(uint8_t a, uint8_t r, uint8_t g, uint8_t b)
   {
   parts.a=a;
   parts.r=r;
   parts.g=g;
   parts.b=b;
   }

To resume, you have an object named Color of 32 bits which contains 4 objects of 8 bits from the structure parts.

Not fully understanding this part of code

2 Answers2