Re: [net-next PATCH] net: phy: aquantia: drop wrong endianness conversion for addr and CRC

Russell King (Oracle)
2023-11-22T22:32:03Z

On Wed, Nov 22, 2023 at 10:09:54PM +0100, Christian Marangi wrote:
> On Wed, Nov 22, 2023 at 08:25:16PM +0000, Russell King (Oracle) wrote:
> > On Wed, Nov 22, 2023 at 08:55:17PM +0100, Christian Marangi wrote:
> > > On Wed, Nov 22, 2023 at 06:53:50PM +0000, Russell King (Oracle) wrote:
> > > > On Wed, Nov 22, 2023 at 06:53:39PM +0100, Christian Marangi wrote:
> > > > > On Wed, Nov 22, 2023 at 05:24:33PM +0000, Russell King (Oracle) wrote:
> > > > > > On Wed, Nov 22, 2023 at 06:08:13PM +0100, Christian Marangi wrote:
> > > > > > > On further testing on BE target with kernel test robot, it was notice
> > > > > > > that the endianness conversion for addr and CRC in fw_load_memory was
> > > > > > > wrong and actually not needed. Values in define doesn't get converted
> > > > > > > and are passed as is and hardcoded values are already in what the PHY
> > > > > > > require, that is LE.
> > > > > > >
> > > > > > > Also drop the cpu_to_be32 for CRC calculation as it's wrong and use
> > > > > > > _swab32 instead, the word is taked from firmware and is always LE, the
> > > > > >
> > > > > > taken
> > > > > >
> > > > > > > mailbox will emit a BE CRC hence the word needs to be always swapped and
> > > > > > > the endianness of the host needs to be ignored.
> > > > > >
> > > > > > I'm not convinced. If the firmware is a bytestream (as most "files" are)
> > > > > > then for val = get_unaligned((u32 *)ptr), where ptr is an array of u8:
> > > > > >
> > > > > > ptr[0] ptr[1] ptr[2] ptr[3] val on LE val on BE
> > > > > > 0x01 0x02 0x03 0x04 0x04030201 0x01020304
> > > > > >
> > > > > > So, endianness matters here, and I think as Jakub already suggested, you
> > > > > > need to use get_unaligned_le32().
> > > > > >
> > > > >
> > > > > So they DO get converted to the HOST endian on reading the firmware from
> > > > > an nvmem cell or a filesystem?
> > > >
> > > > I don't like "converted". It's *not* a conversion. It's a fundamental
> > > > property of accessing memory using different sizes of access.
> > > >
> > > > As I attempted to explain above, if you have a file, and byte 0
> > > > contains 0xAA, byte 1 of the file contains 0xBB, byte 2 contains
> > > > 0xCC, and byte 3 contains 0xDD, then if you read that file byte by
> > > > byte, you will get 0xAA, then 0xBB, then 0xCC and then 0xDD.
> > > >
> > > > If you map that file into memory, e.g. in userspace, using mmap(),
> > > > or allocating memory and reading four bytes into memory, and access
> > > > it using bytes, then at offset 0, you will find 0xAA, offset 1 will
> > > > be 0xBB, etc.
> > > >
> > > > The problems with endianness start when you move away from byte
> > > > access.
> > > >
> > > > If you use 16-bit accessors, then, a little endian machine is defined
> > > > that a 16-bit load from memory will result in the first byte being put
> > > > into the LSB of the 16-bit value, and the second byte will be put into
> > > > the MSB of the 16-bit value. So that would be 0xBBAA. However, on a big
> > > > endian machine, a 16-bit load will result in the first byte being put
> > > > into the MSB of the 16-bit value, and the second byte will be put into
> > > > the LSB of that value - meaning the 16-bit value will be 0xAABB.
> > > >
> > > > The second 16-bit value uses the next two bytes, and the order at which
> > > > these two bytes are placed into the 16-bit value reflects the same as
> > > > the first two bytes. So LE will be 0xDDCC and BE would be 0xCCDD.
> > > >
> > > > The same "swapping" happens with 32-bit, but of course instead of just
> > > > two bytes, it covers four bytes. On LE, a 32-bit access will give
> > > > 0xDDCCBBAA. On BE, that will be 0xAABBCCDD.
> > > >
> > > > Again, this is not to do with any kind of "conversion" happening in
> > > > software. It's a property of how the memory subsystem inside the CPU
> > > > works.
> > > >
> > > > > Again this is really dumping raw data from the read file directly to the
> > > > > mailbox. Unless phy_write does some conversion internally, but in that
> > > > > case how does it know what endian is the PHY internally?
> > > >
> > > > phy_write() does *no* conversion. The MDIO bus defines that a 16-bit
> > > > register value will be transferred, and the MDIO bus specifies that
> > > > bit 15 will be sent first, followed by subsequent bits down to bit 0.
> > > >
> > > > The access to the hardware to make this happen is required to ensure
> > > > that the value passed to phy_write() and read using phy_read() will
> > > > reflect this. So, if one does this:
> > > >
> > > > val = phy_read(phydev, 0);
> > > >
> > > > for (i = 15; i >= 0; i--)
> > > > printk("%u", !!(val & BIT(i)));
> > > >
> > > > printk("\n");
> > > >
> > > > This will give you the stream of bits in the _order_ that they appeared
> > > > on the MDIO bus when phy_read() accessed. Doing the same with a value
> > > > to be written will produce the bits in the same value that they will
> > > > be placed on the MDIO bus.
> > > >
> > > > So, this means that if the BMCR contains 0x1234 in the PHY, phy_read()
> > > > will return 0x1234. Passing 0x1234 into phy_write() will write 0x1234
> > > > in that register. The host endian is entirely irrelevant here.
> > > >
> > >
> > > Thanks a lot for the clarification. And sorry for misusing the word
> > > conversion.
> > >
> > > > > > I would make this explicit:
> > > > > >
> > > > > > u8 crc_data[4];
> > > > > >
> > > > > > ...
> > > > > >
> > > > > > /* CRC is calculated using BE order */
> > > > > > crc_data[0] = word >> 24;
> > > > > > crc_data[1] = word >> 16;
> > > > > > crc_data[2] = word >> 8;
> > > > > > crc_data[3] = word;
> > > > > >
> > > > > > crc = crc_ccitt_false(crc, crc_data, sizeof(crc_data));
> > > > > >
> > > > > > which will be (a) completely unambiguous, and (b) completely
> > > > > > independent of the host endianness.
> > > > >
> > > > > But isn't this exactly what is done with ___constant_swab32 ?
> > > > > __swab32 should not change if the HOST is BE or LE.
> > > >
> > > > Let try again to make this clear. If one has this code:
> > > >
> > > > u32 word = 0x01020304;
> > > > u8 *ptr;
> > > > int i;
> > > >
> > > > ptr = (u8 *)&word;
> > > >
> > > > for (i = 0; i < 4; i++)
> > > > printk(" %02x", ptr[i]);
> > > > printk("\n");
> > > >
> > > > Then, on a:
> > > > - LE machine, this will print " 04 03 02 01"
> > > > - BE machine, this will print " 01 02 03 04"
> > > >
> > > > Now, if you look at the definition of crc_ccitt_false(), it is
> > > > defined to do:
> > > >
> > > > while (len--)
> > > > crc = crc_ccitt_false_byte(crc, *buffer++);
> > > >
> > > > So, on a LE machine, this will feed the above bytes in the order of
> > > > 0x04, 0x03, 0x02, 0x01 in a LE machine, and 0x01, 0x02, 0x03, 0x04
> > > > on a BE machine.
> > > >
> > >
> > > So it's really a problem of setting u8 in word and the order they are
> > > read in the system.
> >
> > Correct.
> >
> > > The first get_unaligned has to be changed to get_unaligned_le32 based on
> > > how the data are treated from passing from an u8 to u32.
> >
> > Yes.
> >
> > I'm going to use the term "bytestream", abbreviated to just stream, to
> > represent the firmware that you are going to upload, because that's
> > essentially what all files are.
> >
> > the first byte of the stream to appear in bits 7:0 of
> > VEND1_GLOBAL_MAILBOX_INTERFACE6
> >
> > the second byte of the stream to appear in bits 15:8 of
> > VEND1_GLOBAL_MAILBOX_INTERFACE6
> >
> > the third byte of the stream to appear in bits 7:0 of
> > VEND1_GLOBAL_MAILBOX_INTERFACE5
> >
> > the forth byte of the stream to appear in bits 15:8 of
> > VEND1_GLOBAL_MAILBOX_INTERFACE5
> >
> > and this to repeat over subsequent groups of four bytes in the stream.
> >
> > This will be achieved by reading the stream using 32-bit little endian
> > accesses using get_unaligned_le32(), and then as you are already doing,
> > splitting them up into two 16-bit quantities.
> >
> > > For LE this doesn't matter but for BE they needs to be swapped as this
> > > is what mailbox expect.
> >
> > Correct.
> >
> > > For CRC. Would something like this work?
> > >
> > > Define u8 crc_data[4];
> > >
> > > *crc_data = (__force u32)cpu_to_be32(word);
> >
> > That won't do what you want, it will only write the first byte.
> >
>
> Right I'm stupid...
>
> Just an example, the correct way would have been...
>
> u8 crc_data[4];
> u32 *crc_word;
> u32 word;
>
> crc_word = (u32 *)crc_data;
> *crc_word = (__force u32)cpu_to_be32(word);

So, let's say "word" was originally 0x12345678.

cpu_to_be32() on LE will make this 0x78563412, and that will be stored
in the array as 0x12, 0x34, 0x56, 0x78.

cpu_to_be32() on BE will be a no-op, so this will remain as 0x12345678,
which will be stored in the array as 0x12, 0x34, 0x56, 0x78.

So yes, that looks like it will work, but there's another issue to
consider - whether this is legal C, or whether it is venturing into
undefined behaviour.

https://gist.github.com/shafik/848ae25ee209f698763cffee272a58f8

> > If you are absolutely certain that each group of four bytes in the
> > source bytestream need to be provided to the CRC function in the
> > reverse order to which they appear in the file.

This is a point to which you unfortunately didn't reply, but is an
important one.

If the bytes in the bytestream are in the correct order for calculating
the CRC, then we can avoid all this and just call the CRC function over
the entire image without needing to do the word load and dance about
changing the order of the bytes - and this was why I mentioned this
hoping to prompt you to double-check.

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