I've been working through Frank D. Luna's book "Introduction to 3D programming with DirectX 10" and one of the problems is to switch from using


D3DXCOLOR color (128 bits)
to UINT color (32 bits)

Presumably the format code to use is: DXGI_FORMAT_R8G8B8A8_UNORM.


In my mind this means you have a variable which at the byte level has information about the channels in the exact order: RGBA (Is this the correct interpretation?--Asking because I'm sure I've read that when you want RGBA you really need a code like: A#R#G#B# where the alpha channel is specified first.

Anyway, I opted (there's probably a better way) to do:
UINT color = (UINT)WHITE;

where WHITE is defined: const D3DXCOLOR WHITE(1.0f, 1.0f, 1.0f, 1.0f);

This cast is defined in the extension to D3DXCOLOR.
However, when DXGI_FORMAT_R8G8B8A8_UNORM is used with the UINT color variable you get the wrong results. Luna attributes this to endianness.


Is this because the cast from D3DXCOLOR produces a UINT of the form RGBA but because intel x86 uses little endiann then at byte level you really get 'ABGR'?? So when this variable actually gets interpreted the shader sees ABGR instead of RGBA? Shouldn't it just know when interpreting bytes that the higher order bits are at the smaller address? And the last question: Since the code is specified as DXGI_FORMAT_R8G8B8A8_UNORM, does this mean that R should be the smallest address and A should be at the largest? I'm sure there are a ton of misconceptions I have so please feel free to dispel them.

Ok so just to make sure I'm getting it right:


D3DXCOLOR::operator UINT () const
{
UINT dwR = r >= 1.0f ? 0xff : r <= 0.0f ? 0x00 : (UINT) (r * 255.0f + 0.5f);
UINT dwG = g >= 1.0f ? 0xff : g <= 0.0f ? 0x00 : (UINT) (g * 255.0f + 0.5f);
UINT dwB = b >= 1.0f ? 0xff : b <= 0.0f ? 0x00 : (UINT) (b * 255.0f + 0.5f);
UINT dwA = a >= 1.0f ? 0xff : a <= 0.0f ? 0x00 : (UINT) (a * 255.0f + 0.5f);
return (dwA << 24) | (dwR << 16) | (dwG << 8) | (dwB << 0);
}


the return statement is essentially a 32 bit entity that looks like

| --byte1=Alpha--|--byte2=Red--|--byte3=Green--|--byte4=Blue--|

Does it make sense that if x86 intel machines are little-endian that I should think of this as


| --byte1=Alpha--|--byte2=Red--|--byte3=Green--|--byte4=Blue--|
LOW ------------------------------------------------------------------- HIGH



-----------------------------------------------------------------------------------------
The function that I used was given by Luna as:


D3DX10INLINE UINT ARGB2ABGR(UINT argb)
{
BYTE A = (argb >> 24) & 0xff;
BYTE R = (argb >> 16) & 0xff;
BYTE G = (argb >> 8) & 0xff;
BYTE B = (argb >> 0) & 0xff;
return (A << 24) | (B << 16) | (G << 8) | (R << 0);
}


Which to me seems to imply:

| --byte1=Alpha--|--byte2=Blue--|--byte3=Green--|--byte4=Red--|
LOW -------------------------------------------------------------------- HIGH



This works but obviously my understanding of the endianness is incorrect because the format code I am using is:
DXGI_FORMAT_R8G8B8A8_UNORM

So this means the expected byte order is Red, Green, Blue, Alpha. Which if you read opposite of what I have above makes sense. To me it looks like the bytes are ordered backwards in memory (ABGR). Maybe I'm misunderstanding the shifting operation.

Does: (A << 24) | (B << 16) | (G << 8) | (R << 0) create the following sort of thing


[font=courier new,courier,monospace]AAAAAAAA000000000000000000000000 (A bits shifted 24 to the left) OR'd With
00000000BBBBBBBB0000000000000000 (B bits shifted 16 to the left) OR'd With
0000000000000000GGGGGGGG00000000 (G bits shifted 8 to the left) OR'd With
000000000000000000000000RRRRRRRR (R bits shifted 0 to the left) Which results in
-------------------------------
AAAAAAAABBBBBBBBGGGGGGGGRRRRRRRR[/font]


[font=courier new,courier,monospace]Is it just a case of convention where I should start at the far right and call that byte 1 or am I missing something else?[/font]
Also does the + 0.5f in (r * 255.0f + 0.5f) and the like, cause rounding upwards or is it doing something else?