Re: [Qemu-devel] [PATCH RESEND for 2.3 4/6] xbzrle: check 8 bytes at a time after an concurrency scene

[Juan Quintela]

<arei.gong...@huawei.com> wrote:
> From: ChenLiang <chenlian...@huawei.com>
>
> The logic of old code is correct. But Checking byte by byte will
> consume time after an concurrency scene.
>
> Signed-off-by: ChenLiang <chenlian...@huawei.com>
> Signed-off-by: Gonglei <arei.gong...@huawei.com>
> ---
>  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?


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;
        }
}