On Fri, 5 Oct 2007, Timur Tabi wrote:
> Andreas Schwab wrote:
>
>> The bit mapping on your device is strictly internal to the device and
>> has nothing to do with bit order on the C level.
>
> Then I don't understand that point of defining __LITTLE_ENDIAN_BITFIELD.
What does it mean for a C-level bitfield ordering to be little-endian if the
processor is BIG_ENDIAN?
>
> --
> Timur Tabi
> Linux Kernel Developer @ Freescale
> -
It makes no sense because a bitfield is something having to
do with a 'C' compiler and it must NEVER be used as a template
to address hardware! 'C' gives no guarantee of the ordering
within machine words. The only way you can access them is
using 'C'. They don't have addresses like other objects
(of course they do exist --somewhere). They are put into
"storage units," according to the standard, and these
storage units are otherwise undefined although you can
align them (don't go there).
If you want to call machine-control bits by name, just
define them as hexadecimal numbers (unsigned ints) and,
if your hardware is for both little/big endian, use
a macro that resolves the issue between the number
and the hardware.
Cheers,
Dick Johnson
Penguin : Linux version 2.6.16.24 on an i686 machine (5592.59 BogoMips).
My book : http://www.AbominableFirebug.com/
_
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