On Wed, 2018-07-11 at 14:41 +0800, Jisheng Zhang wrote: > On Tue, 10 Jul 2018 19:19:21 +0300 Andy Shevchenko wrote: > > > +/* > > > + * divisor = div(I) + div(F) > > > + * "I" means integer, "F" means fractional > > > + * quot = div(I) = clk / (16 * baud) > > > + * frac = div(F) * 2^dlf_size > > > + * > > > + * let rem = clk % (16 * baud) > > > + * we have: div(F) * (16 * baud) = rem > > > + * so frac = 2^dlf_size * rem / (16 * baud) = (rem << dlf_size) / > > > (16 > > > * baud) > > > + */ > > > + quot = p->uartclk / (16 * baud); > > > + rem = p->uartclk % (16 * baud); > > > + *frac = DIV_ROUND_CLOSEST(rem << d->dlf_size, 16 * baud > My python coding skill is limited. So I wrote a simple c program to > do the check for common clks and baudrate combination. All passed. I > paste the code here: > OK, I wrote test case in Python: https://gist.github.com/andy-shev/5e980f1d752617ba814725248556ac19 Looks good to me. Please, send v6 and assume my Reviewed-by for entire series. -- Andy Shevchenko <andriy.shevchenko@xxxxxxxxxxxxxxx> Intel Finland Oy -- To unsubscribe from this list: send the line "unsubscribe linux-serial" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
