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

[Thomas Monjalon <thomas@monjalon.net>]

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.
> 
> 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.


> > > 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 :-)