From mboxrd@z Thu Jan 1 00:00:00 1970 From: Andy Shevchenko Subject: Re: [PATCH] spi: Add FSI-attached SPI controller driver Date: Wed, 5 Feb 2020 17:51:52 +0200 Message-ID: References: <1580328504-436-1-git-send-email-eajames@linux.ibm.com> <29f6cc86-69ca-bc88-b6ae-2b1a24c0dae3@linux.vnet.ibm.com> <744f0019-8656-eec1-cb9a-7e70cd042587@linux.ibm.com> Mime-Version: 1.0 Content-Type: text/plain; charset="UTF-8" Cc: Eddie James , linux-spi , Linux Kernel Mailing List , Mark Brown , Joel Stanley , Andrew Jeffery To: Eddie James Return-path: In-Reply-To: <744f0019-8656-eec1-cb9a-7e70cd042587-tEXmvtCZX7AybS5Ee8rs3A@public.gmane.org> Sender: linux-spi-owner-u79uwXL29TY76Z2rM5mHXA@public.gmane.org List-ID: On Tue, Feb 4, 2020 at 6:06 PM Eddie James wrote: > On 2/4/20 5:02 AM, Andy Shevchenko wrote: > > On Mon, Feb 3, 2020 at 10:33 PM Eddie James wrote: > >> On 1/30/20 10:37 AM, Andy Shevchenko wrote: > >>> On Wed, Jan 29, 2020 at 10:09 PM Eddie James wrote: ... > >>>> + struct device *dev; > >>> Isn't fsl->dev the same? > >>> Perhaps kernel doc to explain the difference? > >> > >> No, it's not the same, as dev here is the SPI controller. I'll add a > >> comment. > > Why to have duplication then? > > > Nothing is being duplicated, the two variables are storing entirely > different information, both of which are necessary for each SPI > controller that this driver is driving. Oh, I see now, thanks! ... > >>>> + for (i = 0; i < num_bytes; ++i) > >>>> + rx[i] = (u8)((in >> (8 * ((num_bytes - 1) - i))) & 0xffULL); > >>> Redundant & 0xffULL part. > >>> > >>> Isn't it NIH of get_unalinged_be64 / le64 or something similar? > >> > >> No, these are shift in/out operations. The read register will also have > >> previous operations data in them and must be extracted with only the > >> correct number of bytes. > > Why not to call put_unaligned() how the tail in this case (it's 0 or > > can be easily made to be 0) will affect the result? > > > The shift-in is not the same as any byte-swap or unaligned operation. > For however many bytes we've read, we start at that many bytes > left-shifted in the register and copy out to our buffer, moving right > for each next byte... I don't think there is an existing function for > this operation. For me it looks like u8 tmp[8]; put_unaligned_be64(in, tmp); memcpy(rx, tmp, num_bytes); put_unaligned*() is just a method to unroll the value to the u8 buffer. See, for example, linux/unaligned/be_byteshift.h implementation. > >>>> + return num_bytes; > >>>> +} > >>>> +static int fsi_spi_data_out(u64 *out, const u8 *tx, int len) > >>>> +{ > >>> Ditto as for above function. (put_unaligned ...) > > Ditto. > > > I don't understand how this could work for transfers of less than 8 > bytes, any put_unaligned would access memory that it doesn't own. Ditto. > >>>> +} -- With Best Regards, Andy Shevchenko