Endian/__BYTE_ORDER question

[Joakim Tjernlund]

Wolfgang Denk <wd at denx.de> wrote on 2010/02/11 22:39:00:
>
> Dear Joakim Tjernlund,
>
> In message <OF918AA866.3ED427EB-ONC12576C7.005CBEE4-C12576C7.
> 005CF730 at transmode.se> you wrote:
> >
> > > I have no idea how it is actually done in the kernel code... but gcc
> > > defines it:
> > >
> > > gcc -dM -E -x c - <<<'' | grep ENDIAN
> > > #define __BIG_ENDIAN__ 1
> > > #define _BIG_ENDIAN 1
> >
> > That doesn't define __BYTE_ORDER. Try the same gcc command
> > on a file that #includes <stdlib.h> and you will get both
> > __BIG_ENDIAN and __LITTLE_ENDIAN
>
> For me this appears to work:
>
> On x86:
>
>    $ echo '#include <stdlib.h>' | gcc -dM -E -x c - | grep ENDIAN
>    #define _ENDIAN_H 1
>    #define PDP_ENDIAN __PDP_ENDIAN
>    #define __PDP_ENDIAN 3412
>    #define BIG_ENDIAN __BIG_ENDIAN
>    #define __BYTE_ORDER __LITTLE_ENDIAN
>    #define __LITTLE_ENDIAN 1234
>    #define __BIG_ENDIAN 4321
>    #define LITTLE_ENDIAN __LITTLE_ENDIAN
>
> On PowerPC:
>
>    $ echo '#include <stdlib.h>' | gcc -dM -E -x c - | grep ENDIAN
>    #define __BIG_ENDIAN__ 1
>    #define __PDP_ENDIAN 3412
>    #define __LITTLE_ENDIAN 1234
>    #define BIG_ENDIAN __BIG_ENDIAN
>    #define _BIG_ENDIAN 1
>    #define __BYTE_ORDER __BIG_ENDIAN
>    #define _ENDIAN_H 1
>    #define __BIG_ENDIAN 4321
>    #define PDP_ENDIAN __PDP_ENDIAN
>    #define LITTLE_ENDIAN __LITTLE_ENDIAN
>
> In both cases __BYTE_ORDER is set to a sane value.

Yes, but that was not what I meant. If you look closer you see that both
__BIG_ENDIAN and __LITTLE_ENDIAN are defined in each compiler. This means
you cannot use these two defines directly to tell if your arch is BE or LE.
Guess what the kernel uses :(

I think the kernel should #define __BYTE_ORDER properly so one
can use __BYTE_ORDER in the kernel too.
Just grep for __BYTE_ORDER in the linux tree to see some ugly
workarounds just because __BYTE_ORDER isn't defined.

 Jocke