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Re: [Qemu-devel] [PATCH RESEND for 2.3 4/6] xbzrle: check 8 bytes at a t


From: Dr. David Alan Gilbert
Subject: Re: [Qemu-devel] [PATCH RESEND for 2.3 4/6] xbzrle: check 8 bytes at a time after an concurrency scene
Date: Wed, 10 Dec 2014 10:41:40 +0000
User-agent: Mutt/1.5.23 (2014-03-12)

* Juan Quintela (address@hidden) wrote:
> <address@hidden> wrote:
> > From: ChenLiang <address@hidden>
> >
> > The logic of old code is correct. But Checking byte by byte will
> > consume time after an concurrency scene.
> >
> > Signed-off-by: ChenLiang <address@hidden>
> > Signed-off-by: Gonglei <address@hidden>
> > ---
> >  xbzrle.c | 28 ++++++++++++++++++----------
> >  1 file changed, 18 insertions(+), 10 deletions(-)
> >
> > diff --git a/xbzrle.c b/xbzrle.c
> > index d27a140..0477367 100644
> > --- a/xbzrle.c
> > +++ b/xbzrle.c
> > @@ -50,16 +50,24 @@ int xbzrle_encode_buffer(uint8_t *old_buf, uint8_t 
> > *new_buf, int slen,
> >  
> >          /* word at a time for speed */
> >          if (!res) {
> > -            while (i < slen &&
> > -                   (*(long *)(old_buf + i)) == (*(long *)(new_buf + i))) {
> > -                i += sizeof(long);
> > -                zrun_len += sizeof(long);
> > -            }
> > -
> > -            /* go over the rest */
> > -            while (i < slen && old_buf[i] == new_buf[i]) {
> > -                zrun_len++;
> > -                i++;
> > +            while (i < slen) {
> > +                if ((*(long *)(old_buf + i)) == (*(long *)(new_buf + i))) {
> > +                    i += sizeof(long);
> > +                    zrun_len += sizeof(long);
> > +                } else {
> > +                    /* go over the rest */
> > +                    for (j = 0; j < sizeof(long); j++) {
> > +                        if (old_buf[i] == new_buf[i]) {
> > +                            i++;
> > +                            zrun_len++;
> > +                        } else {
> > +                            break;
> > +                        }
> > +                    }
> > +                    if (j != sizeof(long)) {
> > +                        break;
> > +                    }
> > +                }
> >              }
> >          }
> 
> This still does misaligned reads.  If we want to do aligned stuff,
> something like that looks much better, no?  Notice that where I put
> "break", I mean we have finished, but you get the idea.  Or I am missing 
> something?

You're missing the loop just above this code that's changed that
guarantees alginment by the start of this code.

Dave

> 
> while(i % sizeof(long) != 0) {
>       if (old_buf[i] == new_buf[i]) {
>               i++;
>               zrun_len++;
>       } else {
>               break;
>       }
> }
> 
> while (i < slen) {
>       if ((*(long *)(old_buf + i)) == (*(long *)(new_buf + i))) {
>               i += sizeof(long);
>               zrun_len += sizeof(long);
>       } else {
>               break;
>       }
> }
> 
> for (j = 0; j < sizeof(long); j++) {
>       if (old_buf[i] == new_buf[i]) {
>               i++;
>               zrun_len++;
>       } else {
>               break;
>       }
> }
--
Dr. David Alan Gilbert / address@hidden / Manchester, UK



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