How to figure out the byteorder only with one byte number?

[bernd@petrovitsch.priv.at (Bernd Petrovitsch)]

On Die, 2012-02-21 at 20:30 +0800, Tao Jiang wrote:
[...]
> Now I know in the most modern machine there is no difference between BE and LE
> at so called 'bit order' level.
> Right?

One main difference between *byte* order and *bit* order is:

What are the means to address individual *bits*?
a) Bit shift and masking as in "1 << bit-number":
   This has a mathematical background and - implicitly - the
   least-significant bit has - thus - the number 0.
   I can't even think of an insane reason (let alone a sane one) to
   break the "shift left is for unsigned numbers equivalent to
   doubling" property - apart from the fact that it is defined in that
   way by C - and all other languages I came across. And the same holds
   for all CPUs/assembler instruction sets ....
b) use a bit-field as in "unsigned char b0:1, b1:1, b2:1, b3:1, b4:1,
   b5:1, b6:1, b7:1;":
   It is not defined by any C-standard and is - thus - up to the
   compiler, if b0 == (1 << 0) or b0 == (1 << 7) or anything else.
c) bit-test/st/clr assembler instructions in the architecture: Go read
   *if* your CPU has such stuff and how it relates to the "bit-shift and
   mask" method.
   I would be greatly surprised if it is different (on i386, it is equal
   since ages BTW) mainly because it makes absolutely no sense.
d) There is hardware with bit-addressable memory out there. Go read the
   manual and the same as c)
I doubt that it is different even for really old machines ....

	Bernd
-- 
Bernd Petrovitsch                  Email : bernd@petrovitsch.priv.at
                     LUGA : http://www.luga.at