[S390] __div64_32 for 31 bit.

The clocksource infrastructure introduced with commit
ad596171ed635c51a9eef829187af100cbf8dcf7 broke 31 bit s390.
The reason is that the do_div() primitive for 31 bit always
had a restriction: it could only divide an unsigned 64 bit
integer by an unsigned 31 bit integer. The clocksource code
now uses do_div() with a base value that has the most
significant bit set. The result is that clock->cycle_interval
has a funny value which causes the linux time to jump around
like mad.
The solution is "obvious": implement a proper __div64_32
function for 31 bit s390.

Signed-off-by: Martin Schwidefsky <schwidefsky@de.ibm.com>

1 files changed, 151 insertions, 0 deletions

diff --git a/arch/s390/lib/div64.c b/arch/s390/lib/div64.c
new file mode 100644
index 000000000000..0481f3424a13
--- /dev/null
+++ b/arch/s390/lib/div64.c
@@ -0,0 +1,151 @@
+/*
+ *  arch/s390/lib/div64.c
+ *
+ *  __div64_32 implementation for 31 bit.
+ *
+ *    Copyright (C) IBM Corp. 2006
+ *    Author(s): Martin Schwidefsky (schwidefsky@de.ibm.com),
+ */
+
+#include <linux/types.h>
+#include <linux/module.h>
+
+#ifdef CONFIG_MARCH_G5
+
+/*
+ * Function to divide an unsigned 64 bit integer by an unsigned
+ * 31 bit integer using signed 64/32 bit division.
+ */
+static uint32_t __div64_31(uint64_t *n, uint32_t base)
+{
+	register uint32_t reg2 asm("2");
+	register uint32_t reg3 asm("3");
+	uint32_t *words = (uint32_t *) n;
+	uint32_t tmp;
+
+	/* Special case base==1, remainder = 0, quotient = n */
+	if (base == 1)
+		return 0;
+	/*
+	 * Special case base==0 will cause a fixed point divide exception
+	 * on the dr instruction and may not happen anyway. For the
+	 * following calculation we can assume base > 1. The first
+	 * signed 64 / 32 bit division with an upper half of 0 will
+	 * give the correct upper half of the 64 bit quotient.
+	 */
+	reg2 = 0UL;
+	reg3 = words[0];
+	asm volatile(
+		"	dr	%0,%2\n"
+		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );
+	words[0] = reg3;
+	reg3 = words[1];
+	/*
+	 * To get the lower half of the 64 bit quotient and the 32 bit
+	 * remainder we have to use a little trick. Since we only have
+	 * a signed division the quotient can get too big. To avoid this
+	 * the 64 bit dividend is halved, then the signed division will
+	 * work. Afterwards the quotient and the remainder are doubled.
+	 * If the last bit of the dividend has been one the remainder
+	 * is increased by one then checked against the base. If the
+	 * remainder has overflown subtract base and increase the
+	 * quotient. Simple, no ?
+	 */
+	asm volatile(
+		"	nr	%2,%1\n"
+		"	srdl	%0,1\n"
+		"	dr	%0,%3\n"
+		"	alr	%0,%0\n"
+		"	alr	%1,%1\n"
+		"	alr	%0,%2\n"
+		"	clr	%0,%3\n"
+		"	jl	0f\n"
+		"	slr	%0,%3\n"
+		"	alr	%1,%2\n"
+		"0:\n"
+		: "+d" (reg2), "+d" (reg3), "=d" (tmp)
+		: "d" (base), "2" (1UL) : "cc" );
+	words[1] = reg3;
+	return reg2;
+}
+
+/*
+ * Function to divide an unsigned 64 bit integer by an unsigned
+ * 32 bit integer using the unsigned 64/31 bit division.
+ */
+uint32_t __div64_32(uint64_t *n, uint32_t base)
+{
+	uint32_t r;
+
+	/*
+	 * If the most significant bit of base is set, divide n by
+	 * (base/2). That allows to use 64/31 bit division and gives a
+	 * good approximation of the result: n = (base/2)*q + r. The
+	 * result needs to be corrected with two simple transformations.
+	 * If base is already < 2^31-1 __div64_31 can be used directly.
+	 */
+	r = __div64_31(n, ((signed) base < 0) ? (base/2) : base);
+	if ((signed) base < 0) {
+		uint64_t q = *n;
+		/*
+		 * First transformation:
+		 * n = (base/2)*q + r
+		 *   = ((base/2)*2)*(q/2) + ((q&1) ? (base/2) : 0) + r
+		 * Since r < (base/2), r + (base/2) < base.
+		 * With q1 = (q/2) and r1 = r + ((q&1) ? (base/2) : 0)
+		 * n = ((base/2)*2)*q1 + r1 with r1 < base.
+		 */
+		if (q & 1)
+			r += base/2;
+		q >>= 1;
+		/*
+		 * Second transformation. ((base/2)*2) could have lost the
+		 * last bit.
+		 * n = ((base/2)*2)*q1 + r1
+		 *   = base*q1 - ((base&1) ? q1 : 0) + r1
+		 */
+		if (base & 1) {
+			int64_t rx = r - q;
+			/*
+			 * base is >= 2^31. The worst case for the while
+			 * loop is n=2^64-1 base=2^31+1. That gives a
+			 * maximum for q=(2^64-1)/2^31 = 0x1ffffffff. Since
+			 * base >= 2^31 the loop is finished after a maximum
+			 * of three iterations.
+			 */
+			while (rx < 0) {
+				rx += base;
+				q--;
+			}
+			r = rx;
+		}
+		*n = q;
+	}
+	return r;
+}
+
+#else /* MARCH_G5 */
+
+uint32_t __div64_32(uint64_t *n, uint32_t base)
+{
+	register uint32_t reg2 asm("2");
+	register uint32_t reg3 asm("3");
+	uint32_t *words = (uint32_t *) n;
+
+	reg2 = 0UL;
+	reg3 = words[0];
+	asm volatile(
+		"	dlr	%0,%2\n"
+		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );
+	words[0] = reg3;
+	reg3 = words[1];
+	asm volatile(
+		"	dlr	%0,%2\n"
+		: "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );
+	words[1] = reg3;
+	return reg2;
+}
+
+#endif /* MARCH_G5 */
+
+EXPORT_SYMBOL(__div64_32);