Re: [PATCH v1 11/13] serial: 8250_exar: Use BIT() in exar_ee_read()

[Parker Newman]

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.

Parker