ftrace: Synchronize the initialization of ftrace_opsgrafted

There's some internal state that ftrace_ops needs to have set, but since
it can be declared outside of the ftrace.c code, it calls
ftrace_ops_init() on the ops in every global function. The issue is that
if two tasks call it on the same ops at the same time it is possible to
have the initialization of one corrupt the initialization of the other
call.

Create a ops_mutex to use to synchronize every initialization of the
ftrace_ops. The mutex is taken within checking the ftrace_ops flag that
states it was initializied but the flag is checked again after the mutex
has been taken. Checking first outside the mutex allows it to shortcut
having to take the mutex. But then the check needs to be done again after
the mute is taken in case of races.

Cc: stable@vger.kernel.org
Link: https://patch.msgid.link/20260902095501.6b59af20@gandalf.local.home
Fixes: f04f24fb7e48d ("ftrace, kprobes: Fix a deadlock on ftrace_regex_lock")
Reported-by: sashiko-bot@kernel.org
Close: https://lore.kernel.org/all/20260829025528.49A831F000E9@smtp.kernel.org/
Signed-off-by: Steven Rostedt <rostedt@goodmis.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;
+}