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SubjectRe: [patch 2.6.28-rc9] spi: spi_write_then_read() regression fix
Date
On Sunday 21 December 2008, Linus Torvalds wrote:
>
> On Sat, 20 Dec 2008, David Brownell wrote:
> >
> > All SPI transfers are full duplex, and are packaged as half duplex
> > by either discarding the data that's read ("write only"), or else
> > by writing zeroes ("read only"). That patch wasn't ensuring that
> > zeroes were getting written out during the "half duplex read" part
> > of the transaction; instead, old RX bits were getting sent.
>
> Hmm. In addition, isn't this broken (in that same function):

No -- this is full duplex. The write_then_read() helper is
simplifying a common half-duplex idiom for short operations,
but the harware still does full duplex. Buffer layout is:

Before: WWWWW0000000
After: xxxxxRRRRRRR

That is, for every bit shifted out (W, 0) another one gets
shifted in (x, R). The I/O primitive essentially swaps
contents of a one-word shift register between master and
slave; or, sequences of such words. Words don't need to
be byte-size, though that's a common option.


> memcpy(local_buf, txbuf, n_tx);
> x.tx_buf = local_buf;
> x.rx_buf = local_buf;
>
> /* do the i/o */
> status = spi_sync(spi, &message);
> if (status == 0)
> memcpy(rxbuf, x.rx_buf + n_tx, n_rx);
>
> shouldn't that 'rx_buf' setup be
>
> x.rx_buf = local_buf + n_tx;
>
> since the whole point was that we allocated a buffer that can hold _both_
> the rx and tx parts? Especially as that final copy into the resulting
> "rxbuf" thing uses that "+ n_tx" addition?

See above. We only want the "R" bits which were shifted in
right *after* the n_tx "W" bits. If we offset rx_buf before
the I/O, we'd start with the "x" don't-care bits and need to
do something else to discard them. (Plus, allocate more
space at the end of the buffer.)

- Dave


> Linus
>
>




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