On Tue, Jan 27, 2015 at 10:03:32PM +0000, Al Viro wrote: > On Tue, Jan 27, 2015 at 04:25:16PM +0100, Karl Beldan wrote: > > The carry from the 64->32bits folding was dropped, e.g with: > > saddr=0xFFFFFFFF daddr=0xFF0000FF len=0xFFFF proto=0 sum=1 > > > > Signed-off-by: Karl Beldan <karl.bel...@rivierawaves.com> > > Cc: Mike Frysinger <vap...@gentoo.org> > > Cc: Arnd Bergmann <a...@arndb.de> > > Cc: linux-kernel@vger.kernel.org > > Cc: Stable <sta...@vger.kernel.org> > > --- > > lib/checksum.c | 4 ++-- > > 1 file changed, 2 insertions(+), 2 deletions(-) > > > > diff --git a/lib/checksum.c b/lib/checksum.c > > index 129775e..4b5adf2 100644 > > --- a/lib/checksum.c > > +++ b/lib/checksum.c > > @@ -195,8 +195,8 @@ __wsum csum_tcpudp_nofold(__be32 saddr, __be32 daddr, > > #else > > s += (proto + len) << 8; > > #endif > > - s += (s >> 32); > > - return (__force __wsum)s; > > + s += (s << 32) + (s >> 32); > > + return (__force __wsum)(s >> 32); > > Umm... I _think_ it's correct, but it needs a better commit message. AFAICS, > what we have is that s is guaranteed to be (a << 32) + b, with a being small. > What we want is something congruent to a + b modulo 0xffff. And yes, in case > when a + b >= 2^32, the original variant fails - it yields a + b - 2^32, which > is one less than what's needed. New one results first in > (a + b)(2^32+1)mod 2^64, then that divided by 2^32. If a + b <= 2^32 - 1, > the first product is less than 2^64 and dividing it by 2^32 yields a + b. > If a + b = 2^32 + c, c is guaranteed to be small and we first get > 2^32 * c + 2^32 + 1, then c + 1, i.e. a + b - 0xffffffff, i.e. > a + b - 0x10001 * 0xffff, so the congruence holds in all cases. > > IOW, I think the fix is correct, but it really needs analysis in the commit > message.
My take on this was "somewhat" simpler: s = a31..0b31..b0 = a << 32 + b, as you put it Here however I don't assume that a is "small", however I assume it has never overflowed, which is trivial to verify since we only add 3 32bits values and 2 16 bits values to a 64bits. Now we just want (a + b + carry(a + b)) % 2^32, and here I assume (a + b + carry(a + b)) % 2^32 == (a + b) % 2^32 + carry(a + b), I guess this is the trick, and this is easy to convince oneself with: 0xffffffff + 0xffffffff == 0x1fffffffe ==> ((u32)-1 + (u32)-1 + 1) % 2^32 == 0xfffffffe % 2^32 + 1 We get this carry pushed out from the MSbs side by the s+= addition pushed back in to the LSbs side of the upper 32bits and this carry doesn't make the upper side overflow. If this explanation is acceptable, I can reword the commit message with it. Sorry if my initial commit log lacked details, and thanks for your detailed input. Karl -- To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majord...@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
