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From: Morten Brørup <mb@smartsharesystems.com>
To: "Thomas Monjalon" <thomas@monjalon.net>, <guohongzhi1@huawei.com>
Cc: <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>
Subject: Re: [dpdk-stable] [dpdk-dev]  [PATCH] bugfix: rte_raw_checksum
Date: Wed, 24 Jun 2020 17:11:19 +0200
Message-ID: <98CBD80474FA8B44BF855DF32C47DC35C610BB@smartserver.smartshare.dk> (raw)
In-Reply-To: <7893780.fyxb4ROZLc@thomas>

> 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.
> >
> > 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 <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 :-)
> 
> 


  reply index

Thread overview: 8+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2020-05-27 13:40 [dpdk-stable] " guohongzhi
2020-05-27 14:58 ` [dpdk-stable] [dpdk-dev] " Morten Brørup
2020-06-24 12:21 ` [dpdk-stable] " Thomas Monjalon
2020-06-24 13:00   ` Morten Brørup
2020-06-24 15:04     ` Thomas Monjalon
2020-06-24 15:11       ` Morten Brørup [this message]
2020-07-06  7:36         ` [dpdk-stable] [dpdk-dev] " Olivier Matz
2020-07-06  7:46           ` Olivier Matz

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