Re: [dpdk-stable] [dpdk-dev] [PATCH] bugfix: rte_raw_checksum

[Olivier Matz <olivier.matz@6wind.com>]

On Mon, Jul 06, 2020 at 09:36:25AM +0200, Olivier Matz wrote:
> Hi Hongzhi,
> 
> I suggest the following title instead:
> 
>   net: fix checksum on big endian CPUs
> 
> On Wed, Jun 24, 2020 at 05:11:19PM +0200, Morten Brørup wrote:
> > > From: dev [mailto:dev-bounces@dpdk.org] On Behalf Of Thomas Monjalon
> > > Sent: Wednesday, June 24, 2020 5:04 PM
> > > 
> > > 24/06/2020 15:00, Morten Brørup:
> > > > > From: Thomas Monjalon [mailto:thomas@monjalon.net]
> > > > > Sent: Wednesday, June 24, 2020 2:22 PM
> > > > >
> > > > > 27/05/2020 15:40, guohongzhi:
> > > > > > From: Hongzhi Guo <guohongzhi1@huawei.com>
> > > > > >
> > > > > > __rte_raw_cksum should consider Big Endian.
> > > > >
> > > > > We need to explain the logic in the commit log.
> 
> Here is a suggestion of commit log:
> 
>   With current code, the checksum of odd-length buffers is wrong on
>   big endian CPUs: the last byte is not properly summed to the
>   accumulator.
> 
>   Fix this by left-shifting the remaining byte by 8. For instance,
>   if the last byte is 0x42, we should add 0x4200 to the accumulator
>   on big endian CPUs.
> 
>   This change is similar to what is suggested in Errata 3133 of
>   RFC 1071.
> 
> Can you please submit a new version with the 2 changes above?

One more thing, please also add:

  Fixes: 6006818cfb26 ("net: new checksum functions")
  Cc: stable@dpdk.org

Thanks
Olivier

> 
> > > >
> > > > Having grown up with big endian CPUs, reading the final byte like
> > > this is obvious to me. I struggle understanding the little endian way
> > > of reading the last byte. (Not really anymore, but back when little
> > > endian was unfamiliar to me I would have struggled.)
> > > >
> > > > An RFC (I can't remember which) describes why the same checksum
> > > calculation code works on both big and little endian CPUs. Is it this
> > > explanation you are asking for?
> > > 
> > > This explanation may be interesting.
> > > 
> > 
> > RFC 1071, especially chapter 3.
> > 
> > Please note that big endian is considered "Normal" order in the RFC. :-)
> 
> There is an errata for this RFC about the C code:
> see https://www.rfc-editor.org/errata/eid3133
> 
> > > 
> > > > > > Signed-off-by: Hongzhi Guo <guohongzhi1@huawei.com>
> > > > > > ---
> > > > > > +#if (RTE_BYTE_ORDER == RTE_BIG_ENDIAN)
> > > > > > +		sum += *((const uint8_t *)u16_buf) << 8;
> > > > > > +#else
> > > > > >  		sum += *((const uint8_t *)u16_buf);
> > > > > > +#endif
> > > > >
> > > > > *((const uint8_t *)u16_buf) should be an uint8_t.
> > > > > What is the expected behaviour of shifting 8 bits of a byte?
> > > >
> > > > Yes, the value will be an uint8_t type. But the shift operation will
> > > cause the compiler to promote the type to int before shifting it.
> > > 
> > > This is the explanation I was looking for :-)
> > > 
> > > 
> >