Re: [PATCH 1/1] Crypto: [xp ]cbc: use 64bit regs on 64bit machines

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On Thu, Jun 14, 2007 at 01:47:19PM +0200, Sebastian Siewior (linux-crypto@xxxxxxxxxxxxxxxx) wrote:
> * Evgeniy Polyakov | 2007-06-14 14:57:45 [+0400]:
> 
> >On Thu, Jun 14, 2007 at 12:20:06AM +0200, Sebastian Siewior (linux-crypto@xxxxxxxxxxxxxxxx) wrote:
> >>  static void xor_64(u8 *a, const u8 *b, unsigned int bs)
> >>  {
> >> +#if BITS_PER_LONG == 64
> >> +	((u64 *)a)[0] ^= ((u64 *)b)[0];
> >> +#else
> >>  	((u32 *)a)[0] ^= ((u32 *)b)[0];
> >>  	((u32 *)a)[1] ^= ((u32 *)b)[1];
> >> +#endif
> >>  }
> >
> >What about endianess?
> 
> It doesn't effect the xor operation. Either I load both LE, xor, store
> LE or load both BE, xor, store BE. The result is always the same. An
> additional endian convert would make thinks just slower.
> Since you confused me a little, I tested it on ppc and amd64 :)

Yeah, logic is correct for both endian.

-- 
	Evgeniy Polyakov
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