On Fri, Nov 10, 2006 at 04:06:52PM -0500, Brian Haley wrote:
> Al Viro wrote:
> >On Fri, Nov 10, 2006 at 02:04:32PM -0500, Brian Haley wrote:
> >>Al Viro wrote:
> >>>so -= 1 is broken even on ia64 and it's *always* broken on big-endian
> >>>boxen.
> >>It's not broken in ia64, I've tested that, just don't have an x86 for
> >>testing right now. Can you please apply these changes and prove it's
> >>broken? This little trick has been done in other UNIXes for years
> >>without any problems.
> >
> >Could you fscking read what you've replied to? Your -=1 will turn 0
> >into 0xffff instead of correct 0xfffe. IOW, it's broken in 1:65536
> >cases.
>
> I looked again at your previous email:
>
> >Note that even on little-endian you want
> > 3 -> 2
> > 2 -> 1
> > 1 -> 0xffff
> > 0 -> 0xfffe
>
> That doesn't look right to me, but I'll take your word that there's one
> edge case out there I don't see (even though this worked on Alpha).
Sigh... Here's how the checksum is defined:
take array of bytes
pad it to even length with 0
split it into 16bit words
define x # y as (x + y) < 0x10000 ? x + y: x + y - 0xffff
calculate sum = w0 # w1 # w2 # ..... # wn
take 0xffff - sum (aka ~sum)
store the resulting value in pair of bytes (with whatever
endianness you used all along).
Note that resulting pair of bytes does *not* depend on endianness. See
aforementioned RFC for proof and for more fun properties of that sucker.
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