[i2c] [PATCH] Add a new-style driver for most I2C EEPROMs

Jean Delvare khali at linux-fr.org
Thu Apr 17 18:24:03 CEST 2008


On Thu, 17 Apr 2008 18:05:32 +0200, Wolfram Sang wrote:
> Greetings,
> 
> (this goes mostly to David, I assume)
> 
> In at24_ee_write:
> > +	/* Writes fail if the previous one didn't complete yet.  We'll
> > +	 * loop a few times until this one succeeds, waiting at least
> > +	 * long enough for one entire page write to work.
> > +	 */
> > +	timeout = jiffies + msecs_to_jiffies(AT24_EE_TIMEOUT);
> > +	for (retries = 0; retries < 3; retries++) {
> > +
> > +		if (at24->use_smbus) {
> > +			status = i2c_smbus_write_i2c_block_data(client,
> > +					offset, count, buf);
> > +			if (status == 0)
> > +				status = count;
> > +		} else {
> > +			status = i2c_transfer(client->adapter, &msg, 1);
> > +			if (status == 1)
> > +				status = count;
> > +		}
> > +		dev_dbg(&client->dev, "write %zd@%d --> %zd (%ld)\n",
> > +				count, offset, status, jiffies);
> > +
> > +		if (status == count)
> > +			return count;
> > +
> > +		if (retries < 3 || time_after(timeout, jiffies)) {
> > +			/* REVISIT:  at HZ=100, this is sloooow */
> > +			msleep(1);
> > +			continue;
> > +		}
> > +	}
> I assume 'retries < 3' (always true) and 'continue' (nothing follows) 
> are left-overs from earlier revisions and can be thrown out? My main 
> questions is 'msleep(1)' though: As I understand it, this means the 
> for-loop will wait roughly 3ms on failures until it reports a timeout 
> (assuming good precision of msleep). Comparing this to AT24_EE_TIMEOUT 
> (25 ms), it looks to me that the msleep value could be increased a 
> little (maybe to AT24_EE_TIMEOUT / 3)? Or is i2c_transfer slow enough 
> and can we count on that behaviour?

msleep() can't sleep less than one jiffie, so msleep(1) is the same as
msleep(10) at HZ=100. At HZ=1000, msleep(1) is almost possible,
although it might sleep for up to 2 ms in practice (finish the current
jiffie + the next jiffie.)

The bottom line is that you never really know how much msleep(N)
will sleep for small values of N.

I realize that it doesn't really answer your question, but that's still
worth keeping in mind.

-- 
Jean Delvare



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