lists.openwall.net   lists  /  announce  owl-users  owl-dev  john-users  john-dev  passwdqc-users  yescrypt  popa3d-users  /  oss-security  kernel-hardening  musl  sabotage  tlsify  passwords  /  crypt-dev  xvendor  /  Bugtraq  Full-Disclosure  linux-kernel  linux-netdev  linux-ext4  linux-hardening  linux-cve-announce  PHC 
Open Source and information security mailing list archives
 
Hash Suite: Windows password security audit tool. GUI, reports in PDF.
[<prev] [next>] [<thread-prev] [thread-next>] [day] [month] [year] [list]
Date: Wed, 22 Nov 2023 20:25:16 +0000
From: "Russell King (Oracle)" <linux@...linux.org.uk>
To: Christian Marangi <ansuelsmth@...il.com>
Cc: Andrew Lunn <andrew@...n.ch>, Heiner Kallweit <hkallweit1@...il.com>,
	"David S. Miller" <davem@...emloft.net>,
	Eric Dumazet <edumazet@...gle.com>,
	Jakub Kicinski <kuba@...nel.org>, Paolo Abeni <pabeni@...hat.com>,
	Robert Marko <robimarko@...il.com>, netdev@...r.kernel.org,
	linux-kernel@...r.kernel.org, kernel test robot <lkp@...el.com>
Subject: Re: [net-next PATCH] net: phy: aquantia: drop wrong endianness
 conversion for addr and CRC

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.

> crc = crc_ccitt_false(crc, crc_data, sizeof(word));

The point of explicitly assigning each byte is to ensure that it's
obvious that we'll get the right result. If we try to write a 32-bit
value, then we're getting right back into the "how does _this_ CPU
map a 32-bit value to indivudual bytes" endianness problem.

The advantage of writing it out as bytes into a u8 array is that from
a code readability point of view, it's all laid out in plain sight
exactly which part of the 32-bit value ends up where and the order in
which the crc function is going to read those bytes - and it is
independent of whatever the endianess of the host architecture.

> Using u8 array should keep the correct order no matter the endian and
> cpu_to_be32 should correctly swap the word if needed. (in a BE HOST data
> should already be in the right order and in LE has to be swapped right?)

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.

-- 
RMK's Patch system: https://www.armlinux.org.uk/developer/patches/
FTTP is here! 80Mbps down 10Mbps up. Decent connectivity at last!

Powered by blists - more mailing lists