On Fri, May 03, 2024 at 02:56:33PM -0400, Parker Newman wrote:
> On Fri, 3 May 2024 18:35:45 +0300
> Andy Shevchenko <andriy.shevchenko@xxxxxxxxxxxxxxx> wrote:
> > On Fri, May 03, 2024 at 10:26:32AM -0400, Parker Newman wrote:
> > > On Thu, 2 May 2024 20:20:01 +0300
> > > Andy Shevchenko <andriy.shevchenko@xxxxxxxxxxxxxxx> wrote:
> > > > On Thu, May 02, 2024 at 07:08:21PM +0300, Ilpo Järvinen wrote:
> > > > > On Thu, 2 May 2024, Andy Shevchenko wrote:
...
> > > > > > // Send address to read from
> > > > > > - for (i = 1 << (UART_EXAR_REGB_EE_ADDR_SIZE - 1); i; i >>= 1)
> > > > > > - exar_ee_write_bit(priv, (ee_addr & i));
> > > > > > + for (i = UART_EXAR_REGB_EE_ADDR_SIZE - 1; i >= 0; i--)
> > > > > > + exar_ee_write_bit(priv, ee_addr & BIT(i));
> > > > > >
> > > > > > // Read data 1 bit at a time
> > > > > > for (i = 0; i <= UART_EXAR_REGB_EE_DATA_SIZE; i++) {
> > > > > > - data <<= 1;
> > > > > > - data |= exar_ee_read_bit(priv);
> > > > > > + if (exar_ee_read_bit(priv))
> > > > > > + data |= BIT(i);
> > > > >
> > > > > Does this end up reversing the order of bits? In the original, data was left
> > > > > shifted which moved the existing bits and added the lsb but the replacement
> > > > > adds highest bit on each iteration?
> > > >
> > > > Oh, seems a good catch!
> > > >
> > > > I was also wondering, but missed this somehow. Seems the EEPROM is in BE mode,
> > > > so two loops has to be aligned.
> > > >
> > >
> > > I just tested this and Ilpo is correct, the read loop portion is backwards as
> > > expected. This is the corrected loop:
> > >
> > > // Read data 1 bit at a time
> > > for (i = UART_EXAR_REGB_EE_DATA_SIZE; i >= 0; i--) {
> > > if (exar_ee_read_bit(priv))
> > > data |= BIT(i);
> > > }
> > >
> > > I know this looks wrong because its looping from 16->0 rather than the
> > > more intuitive 15->0 for a 16bit value. This is actually correct however
> > > because according to the AT93C46D datasheet there is always dummy 0 bit
> > > before the actual 16 bits of data.
> > >
> > > I hope that helps,
> >
> > Yes, it helps and means that we need that comment to be added to the code. Is
> > it the same applicable to the write part above (for address)? Because AFAIU
> > mine is one bit longer than yours. Maybe in the original code is a bug? Have
> > you tried to read high addresses?
>
> The address portion is 6 bits, nothing extra, so what you have is correct.
>
> The original code was legacy, I cleaned it up but didn't change those loops.
>
> Your method works out the the same number of bits as the legacy method
> which sets bit 5 and has to shift right 6 times to get i = 0 which ends the loop.
Okay, thank your for confirming the correctness of a new code!
--
With Best Regards,
Andy Shevchenko
