From mboxrd@z Thu Jan 1 00:00:00 1970 From: Eddie James Subject: Re: [PATCH] spi: Add FSI-attached SPI controller driver Date: Fri, 7 Feb 2020 13:28:40 -0600 Message-ID: <90973143-bd0a-33cf-9eb8-a83be1a9b415@linux.vnet.ibm.com> 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; format=flowed Content-Transfer-Encoding: 7bit Cc: linux-spi , Linux Kernel Mailing List , Mark Brown , Joel Stanley , Andrew Jeffery To: Andy Shevchenko , Eddie James Return-path: In-Reply-To: Content-Language: en-US Sender: linux-spi-owner-u79uwXL29TY76Z2rM5mHXA@public.gmane.org List-ID: On 2/5/20 9:51 AM, Andy Shevchenko wrote: > 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. Unforunately it is not the same. put_unaligned_be64 will take the highest 8 bits (0xff00000000000000) and move it into tmp[0]. Then 0x00ff000000000000 into tmp[1], etc. This is only correct for this driver IF my transfer is 8 bytes. If, for example, I transfer 5 bytes, then I need 0x000000ff00000000 into tmp[0], 0x00000000ff000000 into tmp[1], etc. So I think my current implementation is correct. Thanks, Eddie > >>>>>> + 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. > >>>>>> +}