> From: dev [mailto:dev-boun...@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:tho...@monjalon.net] > > > Sent: Wednesday, June 24, 2020 2:22 PM > > > > > > 27/05/2020 15:40, guohongzhi: > > > > From: Hongzhi Guo <guohongz...@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. >
RFC 1071, especially chapter 3. Please note that big endian is considered "Normal" order in the RFC. :-) > > > > > Signed-off-by: Hongzhi Guo <guohongz...@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 :-) > >
