  `If the input is out of the range of the allowed values, either largerthan the largest value or closer to zero than the smallest non-zeroallowed value, then a division by zero would occur.In the case of input too large, the division by zero will occur on thefirst iteration.  The best result (largest allowed value) will be foundby always choosing the semi-convergent and excluding the denominatorbased limit when finding it.In the case of the input too small, the division by zero will occur onthe second iteration.  The numerator based semi-convergent should not becalculated to avoid the division by zero.  But the semi-convergent vsprevious convergent test is still needed, which effectively choosesbetween 0 (the previous convergent) vs the smallest allowed fraction(best semi-convergent) as the result.Reported-by: Yiyuan Guo <yguoaz@gmail.com>Signed-off-by: Trent Piepho <tpiepho@gmail.com>--- lib/math/rational.c | 16 +++++++++++----- 1 file changed, 11 insertions(+), 5 deletions(-)diff --git a/lib/math/rational.c b/lib/math/rational.cindex 9781d521963d..c0ab51d8fbb9 100644--- a/lib/math/rational.c+++ b/lib/math/rational.c@@ -12,6 +12,7 @@ #include <linux/compiler.h> #include <linux/export.h> #include <linux/minmax.h>+#include <linux/limits.h>  /*  * calculate best rational approximation for a given fraction@@ -78,13 +79,18 @@ void rational_best_approximation( 		 * found below as 't'. 		 */ 		if ((n2 > max_numerator) || (d2 > max_denominator)) {-			unsigned long t = min((max_numerator - n0) / n1,-					      (max_denominator - d0) / d1);+			unsigned long t = ULONG_MAX; -			/* This tests if the semi-convergent is closer-			 * than the previous convergent.+			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 (2u * t > a || (2u * t == a && d0 * dp > d1 * d)) {+			if (!d1 || 2u * t > a || (2u * t == a && d0 * dp > d1 * d)) { 				n1 = n0 + t * n1; 				d1 = d0 + t * d1; 			}-- 2.26.2`   