PATCH: lm92: port for 2.6 kernel
jchapman at katalix.com
Mon Feb 28 23:44:47 CET 2005
Mark A. Greer wrote:
> Jean Delvare wrote:
>>> Don't the i2c_smbus_[read|write]_word_data() routines pass the u16
>>> data in little endian format because that's the endian of the SMbus?
>> Not exactly. The bytes arrive LSB first, but the i2c bus drivers then
>> convert the word to the endianess of the system. At least they are
>> supposed to, those who don't are broken, if any.
>>> Surely if swab16() is used rather than cpu_to_le16(), different values
>>> are passed to the xxx_word_data() routines depending on cpu endian?
>> No, since the conversion is done by the bus drivers. The only reason why
>> swab16 is used is because the chips do not respect the SMBus specs (with
>> good reasons, but still...)
> Uhh oh... The i2c-mv64xxx.c driver--note that the ctlr is *not* and
> smbus ctlr and only provides one byte at a time to the driver--that I
> wrote simply puts the incoming bytes into the buffer in the order that
> they arrive (and sends them out in the order that they appear in the
> buffer (i.e., as a byte-stream)). That is wrong?!?!
This was what was confusing me too. As I understand it now (Jean please
correct me if I'm wrong) the core code behind the smbus read/write
word_data() routines presents the u16 value as a little endian value.
If the chip driver knows the chip implements 16-bit registers as big
endian then it must swap the bytes, regardless of cpu endian.
I don't think Jean's statement that "the conversion is done by the
bus drivers" is true, but maybe I still don't understand... :)
> So, after the read transfer is completed, I should go thru the buffer
> applying le16_to_cpu() to each pair of bytes? What if there are an odd
> number of bytes? Also, I guess I should apply cpu_to_le16() to the
> buffer before I send it out?
I don't see how a bus driver can do this when all it has is a single
"xfer" routine to transfer a byte stream. I already have a driver
that reads 8 bytes from consecutive byte-wide registers as an 8-byte
transfer. Just because the byte stream has more than one byte, we can't
endian-convert byte pairs...
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