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.
--
With Best Regards,
Andy Shevchenko
