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authorMartin Schwidefsky <schwidefsky@de.ibm.com>2006-09-28 10:55:39 -0400
committerMartin Schwidefsky <schwidefsky@de.ibm.com>2006-09-28 10:55:39 -0400
commitd9f7a745d55527d0d41684b22506a86c4381f7f1 (patch)
treeea8870ef06c3723ad59b78aac97bfe8152894c72 /arch/s390/lib/div64.c
parent1fce518e8e7de62597c823d6d795cafc694e7910 (diff)
[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>
Diffstat (limited to 'arch/s390/lib/div64.c')
-rw-r--r--arch/s390/lib/div64.c151
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
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+++ b/arch/s390/lib/div64.c
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1/*
2 * arch/s390/lib/div64.c
3 *
4 * __div64_32 implementation for 31 bit.
5 *
6 * Copyright (C) IBM Corp. 2006
7 * Author(s): Martin Schwidefsky (schwidefsky@de.ibm.com),
8 */
9
10#include <linux/types.h>
11#include <linux/module.h>
12
13#ifdef CONFIG_MARCH_G5
14
15/*
16 * Function to divide an unsigned 64 bit integer by an unsigned
17 * 31 bit integer using signed 64/32 bit division.
18 */
19static uint32_t __div64_31(uint64_t *n, uint32_t base)
20{
21 register uint32_t reg2 asm("2");
22 register uint32_t reg3 asm("3");
23 uint32_t *words = (uint32_t *) n;
24 uint32_t tmp;
25
26 /* Special case base==1, remainder = 0, quotient = n */
27 if (base == 1)
28 return 0;
29 /*
30 * Special case base==0 will cause a fixed point divide exception
31 * on the dr instruction and may not happen anyway. For the
32 * following calculation we can assume base > 1. The first
33 * signed 64 / 32 bit division with an upper half of 0 will
34 * give the correct upper half of the 64 bit quotient.
35 */
36 reg2 = 0UL;
37 reg3 = words[0];
38 asm volatile(
39 " dr %0,%2\n"
40 : "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );
41 words[0] = reg3;
42 reg3 = words[1];
43 /*
44 * To get the lower half of the 64 bit quotient and the 32 bit
45 * remainder we have to use a little trick. Since we only have
46 * a signed division the quotient can get too big. To avoid this
47 * the 64 bit dividend is halved, then the signed division will
48 * work. Afterwards the quotient and the remainder are doubled.
49 * If the last bit of the dividend has been one the remainder
50 * is increased by one then checked against the base. If the
51 * remainder has overflown subtract base and increase the
52 * quotient. Simple, no ?
53 */
54 asm volatile(
55 " nr %2,%1\n"
56 " srdl %0,1\n"
57 " dr %0,%3\n"
58 " alr %0,%0\n"
59 " alr %1,%1\n"
60 " alr %0,%2\n"
61 " clr %0,%3\n"
62 " jl 0f\n"
63 " slr %0,%3\n"
64 " alr %1,%2\n"
65 "0:\n"
66 : "+d" (reg2), "+d" (reg3), "=d" (tmp)
67 : "d" (base), "2" (1UL) : "cc" );
68 words[1] = reg3;
69 return reg2;
70}
71
72/*
73 * Function to divide an unsigned 64 bit integer by an unsigned
74 * 32 bit integer using the unsigned 64/31 bit division.
75 */
76uint32_t __div64_32(uint64_t *n, uint32_t base)
77{
78 uint32_t r;
79
80 /*
81 * If the most significant bit of base is set, divide n by
82 * (base/2). That allows to use 64/31 bit division and gives a
83 * good approximation of the result: n = (base/2)*q + r. The
84 * result needs to be corrected with two simple transformations.
85 * If base is already < 2^31-1 __div64_31 can be used directly.
86 */
87 r = __div64_31(n, ((signed) base < 0) ? (base/2) : base);
88 if ((signed) base < 0) {
89 uint64_t q = *n;
90 /*
91 * First transformation:
92 * n = (base/2)*q + r
93 * = ((base/2)*2)*(q/2) + ((q&1) ? (base/2) : 0) + r
94 * Since r < (base/2), r + (base/2) < base.
95 * With q1 = (q/2) and r1 = r + ((q&1) ? (base/2) : 0)
96 * n = ((base/2)*2)*q1 + r1 with r1 < base.
97 */
98 if (q & 1)
99 r += base/2;
100 q >>= 1;
101 /*
102 * Second transformation. ((base/2)*2) could have lost the
103 * last bit.
104 * n = ((base/2)*2)*q1 + r1
105 * = base*q1 - ((base&1) ? q1 : 0) + r1
106 */
107 if (base & 1) {
108 int64_t rx = r - q;
109 /*
110 * base is >= 2^31. The worst case for the while
111 * loop is n=2^64-1 base=2^31+1. That gives a
112 * maximum for q=(2^64-1)/2^31 = 0x1ffffffff. Since
113 * base >= 2^31 the loop is finished after a maximum
114 * of three iterations.
115 */
116 while (rx < 0) {
117 rx += base;
118 q--;
119 }
120 r = rx;
121 }
122 *n = q;
123 }
124 return r;
125}
126
127#else /* MARCH_G5 */
128
129uint32_t __div64_32(uint64_t *n, uint32_t base)
130{
131 register uint32_t reg2 asm("2");
132 register uint32_t reg3 asm("3");
133 uint32_t *words = (uint32_t *) n;
134
135 reg2 = 0UL;
136 reg3 = words[0];
137 asm volatile(
138 " dlr %0,%2\n"
139 : "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );
140 words[0] = reg3;
141 reg3 = words[1];
142 asm volatile(
143 " dlr %0,%2\n"
144 : "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" );
145 words[1] = reg3;
146 return reg2;
147}
148
149#endif /* MARCH_G5 */
150
151EXPORT_SYMBOL(__div64_32);