Re: block-sha1: improve code on large-register-set machines

From: Nicolas Pitre <nico@cam.org>
Date: Aug 11, 2009, 19:31 UTC
Message-ID: <alpine.LFD.2.00.0908111517390.10633@xanadu.home>
In-Reply-To: <alpine.LFD.2.00.0908111254290.10633@xanadu.home>

On Tue, 11 Aug 2009, Nicolas Pitre wrote:

Show 14 quoted lines

> On Tue, 11 Aug 2009, Linus Torvalds wrote:
> 
> > On Tue, 11 Aug 2009, Nicolas Pitre wrote:
> > > 
> > > #define SHA_SRC(t) \
> > >   ({ unsigned char *__d = (unsigned char *)&data[t]; \
> > >      (__d[0] << 24) | (__d[1] << 16) | (__d[2] << 8) | (__d[3] << 0); })
> > > 
> > > And this provides the exact same performance as the ntohl() based 
> > > version (4.980s) except that this now cope with unaligned buffers too.
> > 
> > The actual object SHA1 calculations are likely not aligned (we do that 
> > object header thing), and if you can't do the htonl() any better way I 
> > guess the byte-based thing is the way to go..

OK, even better: 4.400s.

This is with this instead of the above:

#define SHA_SRC(t) \
   ({   unsigned char *__d = (unsigned char *)data; \
        (__d[(t)*4 + 0] << 24) | (__d[(t)*4 + 1] << 16) | \
        (__d[(t)*4 + 2] <<  8) | (__d[(t)*4 + 3] <<  0); })

The previous version would allocate a new register for __d and then index it with an offset of 0, 1, 2 or 3. This version always uses the register containing the data pointer with absolute offsets. The binary is a bit smaller too.

Nicolas