dmaengine: fix use-after-free in dma_chan_put() and dma_release_channel()grafted

When dma_device_put() drops the last reference on chan->device->ref,
dma_device_release() runs and may free the dma_device along with its
channels.

dma_chan_put() then still reads chan->device->owner via
dma_chan_to_owner() for the trailing module_put(). KASAN catches it:

	slab-use-after-free in dma_chan_put+0x3e6/0x4c0
	Read of size 8 by task insmod/6319
	Freed by task 6319:
	  kfree+0x225/0x470
	  dma_chan_put+0x395/0x4c0
	  dmaengine_put+0xf8/0x160

Cache the module owner in dma_chan_put() before the put so the trailing
module_put() does not need chan->device.

Fixes: 8ad342a86359 ("dmaengine: Add reference counting to dma_device struct")
Suggested-by: Sashiko <sashiko-bot@kernel.org>
Link: https://sashiko.dev/#/patchset/20260518-dmaengine-kref-fix-v1-1-4d6125048fb7@amd.com
Reviewed-by: Frank Li <Frank.Li@nxp.com>
Reviewed-by: Logan Gunthorpe <logang@deltatee.com>
Signed-off-by: Shivank Garg <shivankg@amd.com>
Link: https://patch.msgid.link/20260822-dmaengine-kref-fix-v5-3-d4a4ee47d927@amd.com
Signed-off-by: Vinod Koul <vkoul@kernel.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");