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, 112 insertions, 0 deletions

diff --git a/lib/math/rational.c b/lib/math/rational.c
new file mode 100644
index 000000000..d2c34e629
--- /dev/null
+++ b/lib/math/rational.c
@@ -0,0 +1,112 @@
+// SPDX-License-Identifier: GPL-2.0
+/*
+ * rational fractions
+ *
+ * Copyright (C) 2009 emlix GmbH, Oskar Schirmer <oskar@scara.com>
+ * Copyright (C) 2019 Trent Piepho <tpiepho@gmail.com>
+ *
+ * helper functions when coping with rational numbers
+ */
+
+#include <linux/rational.h>
+#include <linux/compiler.h>
+#include <linux/export.h>
+#include <linux/minmax.h>
+#include <linux/limits.h>
+#include <linux/module.h>
+
+/*
+ * calculate best rational approximation for a given fraction
+ * taking into account restricted register size, e.g. to find
+ * appropriate values for a pll with 5 bit denominator and
+ * 8 bit numerator register fields, trying to set up with a
+ * frequency ratio of 3.1415, one would say:
+ *
+ * rational_best_approximation(31415, 10000,
+ *		(1 << 8) - 1, (1 << 5) - 1, &n, &d);
+ *
+ * you may look at given_numerator as a fixed point number,
+ * with the fractional part size described in given_denominator.
+ *
+ * for theoretical background, see:
+ * https://en.wikipedia.org/wiki/Continued_fraction
+ */
+
+void rational_best_approximation(
+	unsigned long given_numerator, unsigned long given_denominator,
+	unsigned long max_numerator, unsigned long max_denominator,
+	unsigned long *best_numerator, unsigned long *best_denominator)
+{
+	/* n/d is the starting rational, which is continually
+	 * decreased each iteration using the Euclidean algorithm.
+	 *
+	 * dp is the value of d from the prior iteration.
+	 *
+	 * n2/d2, n1/d1, and n0/d0 are our successively more accurate
+	 * approximations of the rational.  They are, respectively,
+	 * the current, previous, and two prior iterations of it.
+	 *
+	 * a is current term of the continued fraction.
+	 */
+	unsigned long n, d, n0, d0, n1, d1, n2, d2;
+	n = given_numerator;
+	d = given_denominator;
+	n0 = d1 = 0;
+	n1 = d0 = 1;
+
+	for (;;) {
+		unsigned long dp, a;
+
+		if (d == 0)
+			break;
+		/* Find next term in continued fraction, 'a', via
+		 * Euclidean algorithm.
+		 */
+		dp = d;
+		a = n / d;
+		d = n % d;
+		n = dp;
+
+		/* Calculate the current rational approximation (aka
+		 * convergent), n2/d2, using the term just found and
+		 * the two prior approximations.
+		 */
+		n2 = n0 + a * n1;
+		d2 = d0 + a * d1;
+
+		/* If the current convergent exceeds the maxes, then
+		 * return either the previous convergent or the
+		 * largest semi-convergent, the final term of which is
+		 * found below as 't'.
+		 */
+		if ((n2 > max_numerator) || (d2 > max_denominator)) {
+			unsigned long t = ULONG_MAX;
+
+			if (d1)
+				t = (max_denominator - d0) / d1;
+			if (n1)
+				t = min(t, (max_numerator - n0) / n1);
+
+			/* This tests if the semi-convergent is closer than the previous
+			 * convergent.  If d1 is zero there is no previous convergent as this
+			 * is the 1st iteration, so always choose the semi-convergent.
+			 */
+			if (!d1 || 2u * t > a || (2u * t == a && d0 * dp > d1 * d)) {
+				n1 = n0 + t * n1;
+				d1 = d0 + t * d1;
+			}
+			break;
+		}
+		n0 = n1;
+		n1 = n2;
+		d0 = d1;
+		d1 = d2;
+	}
+	*best_numerator = n1;
+	*best_denominator = d1;
+}
+
+EXPORT_SYMBOL(rational_best_approximation);
+
+MODULE_DESCRIPTION("Rational fraction support library");
+MODULE_LICENSE("GPL v2");