On Tue, Feb 07, 2017 at 10:43:13PM -0800, Sudarsana Kalluru wrote:
> +/* Adjust the HW clock by a rate given in parts-per-billion (ppb) units.
> + * FW/HW accepts the adjustment value in terms of 3 parameters:
> + * Drift period - adjustment happens once in certain number of nano
> seconds.
> + * Drift value - time is adjusted by a certain value, for example by 5 ns.
> + * Drift direction - add or subtract the adjustment value.
> + * The routine translates ppb into the adjustment triplet in an optimal
> manner.
> + */
> +static int qed_ptp_hw_adjfreq(struct qed_dev *cdev, s32 ppb)
> +{
> + struct qed_hwfn *p_hwfn = QED_LEADING_HWFN(cdev);
> + s64 period, period1, period2, dif, dif1, dif2;
> + struct qed_ptt *p_ptt = p_hwfn->p_ptp_ptt;
> + int drift_dir, best_val, best_period;
> + s64 best_dif, temp, val;
> + u32 drift_ctr_cfg = 0;
> + u32 drift_state;
> +
> + best_dif = ppb;
> + best_period = 2;
> + best_val = 0;
> + drift_dir = 1;
> +
> + if (ppb < 0) {
> + ppb = -ppb;
> + drift_dir = 0;
> + }
> +
> + if (ppb == 0) {
> + /* No clock adjustment required */
> + best_val = 0;
> + best_period = 0xFFFFFFF;
> + } else {
> + /* Adjustment value is up to +/-7ns, find an optimal value in
> + * this range.
> + */
> + for (val = 1; val <= 7; val++) {
> + period1 = div_s64(val * 1000000000, ppb);
You don't round this, but you should.
> + period1 -= 8;
> + period1 >>= 4;
You should round here as well and spare yourself the following two
divisions.
> + if (period1 < 1)
> + period1 = 1;
> + if (period1 > 0xFFFFFFE)
> + period1 = 0xFFFFFFE;
> + period2 = period1 + 1;
> +
> + temp = div_s64(val * 1000000000, (period1 * 16 + 8));
> + dif1 = ppb - temp;
> + if (dif1 < 0)
> + dif1 = -dif1;
> +
> + temp = div_s64(val * 1000000000, (period2 * 16 + 8));
> + dif2 = ppb - temp;
> + if (dif2 < 0)
> + dif2 = -dif2;
> +
> + dif = min_t(s64, dif1, dif2);
> + period = (dif1 < dif2) ? period1 : period2;
> + if (dif < best_dif) {
Consider testing for (dif <= best) instead.
If I am not mistaken, then you can skip the cases val==2 and val==3,
because they are equivalent to val==4 and 6.
> + best_dif = dif;
> + best_val = (int)val;
> + best_period = (int)period;
> + }
> + }
> + }
Thanks,
Richard
