RE: [tip:x86/core 1/1] arch/x86/um/../lib/csum-partial_64.c:98:12: error: implicit declaration of function 'load_unaligned_zeropad'

David Laight
2021-11-26T17:20:44Z

From: Eric Dumazet
> Sent: 25 November 2021 04:01
...
> > The outputs seem to match if `buff` is aligned to 64-bit. Still see
> > difference with `csum_fold(csum_partial())` if `buff` is not 64-bit aligned.
> >
> > The comment at the top says it's "best" to have `buff` 64-bit aligned but
> > the code logic seems meant to support the misaligned case so not
> > sure if it's an issue.
> >
>
> It is an issue in general, not in standard cases because network
> headers are aligned.
>
> I think it came when I folded csum_partial() and do_csum(), I forgot
> to ror() the seed.
>
> I suspect the following would help:
>
> diff --git a/arch/x86/lib/csum-partial_64.c b/arch/x86/lib/csum-partial_64.c
> index 1eb8f2d11f7c785be624eba315fe9ca7989fd56d..ee7b0e7a6055bcbef42d22f7e1d8f52ddbd6be6d
> 100644
> --- a/arch/x86/lib/csum-partial_64.c
> +++ b/arch/x86/lib/csum-partial_64.c
> @@ -41,6 +41,7 @@ __wsum csum_partial(const void *buff, int len, __wsum sum)
> if (unlikely(odd)) {
> if (unlikely(len == 0))
> return sum;
> + temp64 = ror32((__force u64)sum, 8);
> temp64 += (*(unsigned char *)buff << 8);
> len--;
> buff++;

You can save an instruction (as if this path matters) by:
temp64 = sum + *(unsigned char *)buff;
temp64 <<= 8;
Although that probably falls foul of 64bit shifts being slow.
So maybe just:
sum += *(unsigned char *)buff;
temp64 = bswap32(sum);
AFAICT (from a pdf) bswap32() and ror(x, 8) are likely to be
the same speed but may use different execution units.

Intel seem so have managed to slow down ror(x, %cl) to 3 clocks
in sandy bridge - and still not fixed it.
Although the compiler might be making a pigs-breakfast of the
register allocation when you tried setting 'odd = 8'.

Weeks can be spent fiddling with this code :-(

David

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