scftorture: Count single_rpc offline failures in statistics outputgrafted

scf_torture_stats_print() aggregates each invoker thread's counters
into a local scf_statistics structure before printing, but the
n_single_rpc_ofl field is missing from the aggregation loop.  As a
result, the "single_rpc_ofl" value printed in the statistics line is
always zero, even when smp_call_function_single() invocations for the
RPC test have failed due to offline CPUs and been counted by the
invoker threads.

Add the missing accumulation so that the printed value reflects the
actual counts.

Signed-off-by: Joel Fernandes <joelagnelf@nvidia.com>
Signed-off-by: Paul E. McKenney <paulmck@kernel.org>

1 files changed, 87 insertions, 0 deletions

diff --git a/tools/perf/util/levenshtein.c b/tools/perf/util/levenshtein.c
new file mode 100644
index 000000000..6a6712635
--- /dev/null
+++ b/tools/perf/util/levenshtein.c
@@ -0,0 +1,87 @@
+// SPDX-License-Identifier: GPL-2.0
+#include "levenshtein.h"
+#include <errno.h>
+#include <stdlib.h>
+#include <string.h>
+
+/*
+ * This function implements the Damerau-Levenshtein algorithm to
+ * calculate a distance between strings.
+ *
+ * Basically, it says how many letters need to be swapped, substituted,
+ * deleted from, or added to string1, at least, to get string2.
+ *
+ * The idea is to build a distance matrix for the substrings of both
+ * strings.  To avoid a large space complexity, only the last three rows
+ * are kept in memory (if swaps had the same or higher cost as one deletion
+ * plus one insertion, only two rows would be needed).
+ *
+ * At any stage, "i + 1" denotes the length of the current substring of
+ * string1 that the distance is calculated for.
+ *
+ * row2 holds the current row, row1 the previous row (i.e. for the substring
+ * of string1 of length "i"), and row0 the row before that.
+ *
+ * In other words, at the start of the big loop, row2[j + 1] contains the
+ * Damerau-Levenshtein distance between the substring of string1 of length
+ * "i" and the substring of string2 of length "j + 1".
+ *
+ * All the big loop does is determine the partial minimum-cost paths.
+ *
+ * It does so by calculating the costs of the path ending in characters
+ * i (in string1) and j (in string2), respectively, given that the last
+ * operation is a substitution, a swap, a deletion, or an insertion.
+ *
+ * This implementation allows the costs to be weighted:
+ *
+ * - w (as in "sWap")
+ * - s (as in "Substitution")
+ * - a (for insertion, AKA "Add")
+ * - d (as in "Deletion")
+ *
+ * Note that this algorithm calculates a distance _iff_ d == a.
+ */
+int levenshtein(const char *string1, const char *string2,
+		int w, int s, int a, int d)
+{
+	int len1 = strlen(string1), len2 = strlen(string2);
+	int *row0 = malloc(sizeof(int) * (len2 + 1));
+	int *row1 = malloc(sizeof(int) * (len2 + 1));
+	int *row2 = malloc(sizeof(int) * (len2 + 1));
+	int i, j;
+
+	for (j = 0; j <= len2; j++)
+		row1[j] = j * a;
+	for (i = 0; i < len1; i++) {
+		int *dummy;
+
+		row2[0] = (i + 1) * d;
+		for (j = 0; j < len2; j++) {
+			/* substitution */
+			row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
+			/* swap */
+			if (i > 0 && j > 0 && string1[i - 1] == string2[j] &&
+					string1[i] == string2[j - 1] &&
+					row2[j + 1] > row0[j - 1] + w)
+				row2[j + 1] = row0[j - 1] + w;
+			/* deletion */
+			if (row2[j + 1] > row1[j + 1] + d)
+				row2[j + 1] = row1[j + 1] + d;
+			/* insertion */
+			if (row2[j + 1] > row2[j] + a)
+				row2[j + 1] = row2[j] + a;
+		}
+
+		dummy = row0;
+		row0 = row1;
+		row1 = row2;
+		row2 = dummy;
+	}
+
+	i = row1[len2];
+	free(row0);
+	free(row1);
+	free(row2);
+
+	return i;
+}