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SubjectRe: TCP Segmentation Offloading (TSO)
From
Hello, 

> > While it would work, this sequence is overkill. Unless I'm mistaken, the
> > only property of bswap which is used in this case is that it swaps even
> > and odd bytes, which can be done by a simple "roll $8,%eax" (or rorl).
> >
> > I believe that bswap is one byte shorter than roll. In any case, using a
> > rotate might be the right thing to do on other architectures.
> >
>
> And again, I think you'll find the rotate faster on at least some x86 cores.

Yeah, I replaced "bswap %eax" with "roll $8,%eax" which would be more
familier to us.

> Better fix is to verify len >=2 before half-word alignment
> test at the beginning of csum_partial. I am not enough of
> an x86 coder to hack this up reliably. :-)

Don't care about the order of checking len and half-word alignment
as both of them have to be checked after all.

Thank you,
Hirokazu Takahashi.


--- linux/arch/i386/lib/checksum.S.BUG Sun Sep 1 17:00:59 2030
+++ linux/arch/i386/lib/checksum.S Thu Sep 5 10:33:31 2030
@@ -126,8 +126,8 @@ csum_partial:
movl 16(%esp),%ecx # Function arg: int len
movl 12(%esp),%esi # Function arg: const unsigned char *buf

- testl $2, %esi
- jnz 30f
+ testl $3, %esi
+ jnz 25f
10:
movl %ecx, %edx
movl %ecx, %ebx
@@ -145,6 +145,20 @@ csum_partial:
lea 2(%esi), %esi
adcl $0, %eax
jmp 10b
+25:
+ testl $1, %esi
+ jz 30f
+ # buf is odd
+ dec %ecx
+ jl 90f
+ roll $8, %eax
+ movzbl (%esi), %ebx
+ shll $8, %ebx
+ addl %ebx, %eax
+ adcl $0, %eax
+ inc %esi
+ testl $2, %esi
+ jz 10b

30: subl $2, %ecx
ja 20b
@@ -211,6 +225,10 @@ csum_partial:
addl %ebx,%eax
adcl $0,%eax
80:
+ testl $1, 12(%esp)
+ jz 90f
+ roll $8, %eax
+90:
popl %ebx
popl %esi
ret
-
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