Re: __LITTLE_ENDIAN vs. __LITTLE_ENDIAN_BITFIELD

[Timur Tabi <timur@freescale.com>]

linux-os (Dick Johnson) wrote:

> 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).

Well, if it doesn't make any sense why do we have __LITTLE_ENDIAN_BITFIELD and 
__BIG_ENDIAN_BITFIELD?  That is, why do we do this:

#if defined(__BIG_ENDIAN_BITFIELD)
	__u8 reserved1		: 2;
	__u8 ili		: 1;
	__u8 reserved2		: 1;
	__u8 sense_key		: 4;
#elif defined(__LITTLE_ENDIAN_BITFIELD)
	__u8 sense_key		: 4;
	__u8 reserved2		: 1;
	__u8 ili		: 1;
	__u8 reserved1		: 2;
#endif

when we can just do this:

#if defined(__BIG_ENDIAN)
	__u8 reserved1		: 2;
	__u8 ili		: 1;
	__u8 reserved2		: 1;
	__u8 sense_key		: 4;
#elif defined(__LITTLE_ENDIAN)
	__u8 sense_key		: 4;
	__u8 reserved2		: 1;
	__u8 ili		: 1;
	__u8 reserved1		: 2;
#endif

> 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.

That wasn't my intention.  I was hoping that __LITTLE_ENDIAN_BITFIELD could be 
used to test bit-endianness, but I guess it can't.

--
Timur Tabi
Linux Kernel Developer @ Freescale
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