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Wed, 24 Jun 2020 11:04:09 -0400 (EDT) From: Thomas Monjalon <thomas@monjalon.net> To: guohongzhi1@huawei.com Cc: Morten =?ISO-8859-1?Q?Br=F8rup?= <mb@smartsharesystems.com>, dev@dpdk.org, stable@dpdk.org, olivier.matz@6wind.com, konstantin.ananyev@intel.com, jiayu.hu@intel.com, ferruh.yigit@intel.com, nicolas.chautru@intel.com, cristian.dumitrescu@intel.com, zhoujingbin@huawei.com, chenchanghu@huawei.com, jerry.lilijun@huawei.com, haifeng.lin@huawei.com Date: Wed, 24 Jun 2020 17:04:07 +0200 Message-ID: <7893780.fyxb4ROZLc@thomas> In-Reply-To: <98CBD80474FA8B44BF855DF32C47DC35C610BA@smartserver.smartshare.dk> References: <20200527134009.19444-1-guohongzhi1@huawei.com> <2108086.oFbrkSXBjQ@thomas> <98CBD80474FA8B44BF855DF32C47DC35C610BA@smartserver.smartshare.dk> MIME-Version: 1.0 Content-Transfer-Encoding: quoted-printable Content-Type: text/plain; charset="iso-8859-1" Subject: Re: [dpdk-stable] [PATCH] bugfix: rte_raw_checksum X-BeenThere: stable@dpdk.org X-Mailman-Version: 2.1.15 Precedence: list List-Id: patches for DPDK stable branches <stable.dpdk.org> List-Unsubscribe: <https://mails.dpdk.org/options/stable>, <mailto:stable-request@dpdk.org?subject=unsubscribe> List-Archive: <http://mails.dpdk.org/archives/stable/> List-Post: <mailto:stable@dpdk.org> List-Help: <mailto:stable-request@dpdk.org?subject=help> List-Subscribe: <https://mails.dpdk.org/listinfo/stable>, <mailto:stable-request@dpdk.org?subject=subscribe> Errors-To: stable-bounces@dpdk.org Sender: "stable" <stable-bounces@dpdk.org> 24/06/2020 15:00, Morten Br=F8rup: > > From: Thomas Monjalon [mailto:thomas@monjalon.net] > > Sent: Wednesday, June 24, 2020 2:22 PM > >=20 > > 27/05/2020 15:40, guohongzhi: > > > From: Hongzhi Guo <guohongzhi1@huawei.com> > > > > > > __rte_raw_cksum should consider Big Endian. > >=20 > > We need to explain the logic in the commit log. >=20 > 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 t= he last byte. (Not really anymore, but back when little endian was unfamili= ar to me I would have struggled.) >=20 > An RFC (I can't remember which) describes why the same checksum calculati= on code works on both big and little endian CPUs. Is it this explanation yo= u are asking for? This explanation may be interesting. > > > Signed-off-by: Hongzhi Guo <guohongzhi1@huawei.com> > > > --- > > > +#if (RTE_BYTE_ORDER =3D=3D RTE_BIG_ENDIAN) > > > + sum +=3D *((const uint8_t *)u16_buf) << 8; > > > +#else > > > sum +=3D *((const uint8_t *)u16_buf); > > > +#endif > >=20 > > *((const uint8_t *)u16_buf) should be an uint8_t. > > What is the expected behaviour of shifting 8 bits of a byte? >=20 > Yes, the value will be an uint8_t type. But the shift operation will caus= e the compiler to promote the type to int before shifting it. This is the explanation I was looking for :-)