On 1.08.2018 03:51, Steven Rostedt wrote:
+static void ksmodel_set_in_range_bining(struct kshark_trace_histo *histo, + size_t n, uint64_t min, uint64_t max, + bool force_in_range) +{ + uint64_t corrected_range, delta_range, range = max - min; + struct kshark_entry *last; + + /* The size of the bin must be >= 1, hence the range must be >= n. */ + if (n == 0 || range < n) + return; + + /* + * If the number of bins changes, allocate memory for the descriptor + * of the model. + */ + if (n != histo->n_bins) { + if (!ksmodel_histo_alloc(histo, n)) { + ksmodel_clear(histo); + return; + } + } + + /* Reset the content of all bins (including overflow bins) to zero. */ + ksmodel_reset_bins(histo, 0, histo->n_bins + 1);Here we could then have: ksmodel_reset_bins(histo, 0, ALLB(histo));+ + if (range % n == 0) { + /* + * The range is multiple of the number of bin and needs no + * adjustment. This is very unlikely to happen but still ... + */ + histo->min = min; + histo->max = max; + histo->bin_size = range / n; + } else { + /* + * The range needs adjustment. The new range will be slightly + * bigger, compared to the requested one. + */ + histo->bin_size = range / n + 1; + corrected_range = histo->bin_size * n; + delta_range = corrected_range - range; + histo->min = min - delta_range / 2; + histo->max = histo->min + corrected_range; + + if (!force_in_range) + return; + + /* + * Make sure that the new range doesn't go outside of the time + * interval of the dataset. + */ + last = histo->data[histo->data_size - 1]; + if (histo->min < histo->data[0]->ts) { + histo->min = histo->data[0]->ts; + histo->max = histo->min + corrected_range; + } else if (histo->max > last->ts) { + histo->max = last->ts; + histo->min = histo->max - corrected_range; + }Hmm, Let's say the range of the data is 0..1,000,001 and we picked a range of 999,999 starting at 0. And there's 1024 buckets. This would have: min = 0; max = 999999; n = 1024; range = 999999; bin_size = 999999 / 1024 + 1 = 977; correct_range = 977 * 1024 = 1000448; delta_rang = 1000448 - 999999 = 449; histo->min = 0 - 449 / 2 = -224; histo->max = -224 + 1000448 = 1000224; Now histo->min (-224) < histo->data[0]->ts (0) so histo->min = 0; histo->max = 0 + 1000448 = 1000448; Thus we get max greater than the data set. Actually, we would always get a range greater than the data set, if the data set itself is not divisible by the bin size. This that a problem?
Hi Steven,In your example you consider the case when we want to visualize the entire data-set. Indeed in this case the true range of the histo will be slightly bigger. This is not a problem.
The "force_in_range" part of the logic in this function deals with another case. Let's stick to your example but say that the current range is from 0 to 10,000. Now if the user hits the "Zoom Out" for a second, this function will be called several hundred times and each call of the function will add its own small negative correction to the value of histo->min (initially 0). At the end we will have a significant part of the graph outside of the data-set.
Thanks! Yordan
-- Steve+ } +} + +/** + * @brief Prepare the bining of the Visualization model. + * @param histo: Input location for the model descriptor. + * @param n: Number of bins. + * @param min: Lower edge of the time-window to be visualized. + * @param max: Upper edge of the time-window to be visualized. + */ +void ksmodel_set_bining(struct kshark_trace_histo *histo, + size_t n, uint64_t min, uint64_t max) +{ + ksmodel_set_in_range_bining(histo, n, min, max, false); +} +
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