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review.evervolv.com / android_kernel_htc_msm8960 / ec33679d78f9d653a44ddba10b5fb824c06330a1 / . / arch / sparc / lib / umul.S
blob: 1f36ae682529449f53c96bb473e6ffa235f7d084 [file] [log] [blame]
Adrian Bunk88278ca2008-05-19 16:53:02 -0700[diff] [blame]1/*
Linus Torvalds1da177e2005-04-16 15:20:36 -0700[diff] [blame]2 * umul.S: This routine was taken from glibc-1.09 and is covered
3 * by the GNU Library General Public License Version 2.
4 */
5
6
7/*
8 * Unsigned multiply. Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the
9 * upper 32 bits of the 64-bit product).
10 *
11 * This code optimizes short (less than 13-bit) multiplies. Short
12 * multiplies require 25 instruction cycles, and long ones require
13 * 45 instruction cycles.
14 *
15 * On return, overflow has occurred (%o1 is not zero) if and only if
16 * the Z condition code is clear, allowing, e.g., the following:
17 *
18 * call .umul
19 * nop
20 * bnz overflow (or tnz)
21 */
22
23 .globl .umul
Al Viro7caaeab2005-09-11 20:14:07 -0700[diff] [blame]24 .globl _Umul
Linus Torvalds1da177e2005-04-16 15:20:36 -0700[diff] [blame]25.umul:
Al Viro7caaeab2005-09-11 20:14:07 -0700[diff] [blame]26_Umul: /* needed for export */
Linus Torvalds1da177e2005-04-16 15:20:36 -0700[diff] [blame]27 or %o0, %o1, %o4
28 mov %o0, %y ! multiplier -> Y
29
30 andncc %o4, 0xfff, %g0 ! test bits 12..31 of *both* args
31 be Lmul_shortway ! if zero, can do it the short way
32 andcc %g0, %g0, %o4 ! zero the partial product and clear N and V
33
34 /*
35 * Long multiply. 32 steps, followed by a final shift step.
36 */
37 mulscc %o4, %o1, %o4 ! 1
38 mulscc %o4, %o1, %o4 ! 2
39 mulscc %o4, %o1, %o4 ! 3
40 mulscc %o4, %o1, %o4 ! 4
41 mulscc %o4, %o1, %o4 ! 5
42 mulscc %o4, %o1, %o4 ! 6
43 mulscc %o4, %o1, %o4 ! 7
44 mulscc %o4, %o1, %o4 ! 8
45 mulscc %o4, %o1, %o4 ! 9
46 mulscc %o4, %o1, %o4 ! 10
47 mulscc %o4, %o1, %o4 ! 11
48 mulscc %o4, %o1, %o4 ! 12
49 mulscc %o4, %o1, %o4 ! 13
50 mulscc %o4, %o1, %o4 ! 14
51 mulscc %o4, %o1, %o4 ! 15
52 mulscc %o4, %o1, %o4 ! 16
53 mulscc %o4, %o1, %o4 ! 17
54 mulscc %o4, %o1, %o4 ! 18
55 mulscc %o4, %o1, %o4 ! 19
56 mulscc %o4, %o1, %o4 ! 20
57 mulscc %o4, %o1, %o4 ! 21
58 mulscc %o4, %o1, %o4 ! 22
59 mulscc %o4, %o1, %o4 ! 23
60 mulscc %o4, %o1, %o4 ! 24
61 mulscc %o4, %o1, %o4 ! 25
62 mulscc %o4, %o1, %o4 ! 26
63 mulscc %o4, %o1, %o4 ! 27
64 mulscc %o4, %o1, %o4 ! 28
65 mulscc %o4, %o1, %o4 ! 29
66 mulscc %o4, %o1, %o4 ! 30
67 mulscc %o4, %o1, %o4 ! 31
68 mulscc %o4, %o1, %o4 ! 32
69 mulscc %o4, %g0, %o4 ! final shift
70
71
72 /*
73 * Normally, with the shift-and-add approach, if both numbers are
74 * positive you get the correct result. With 32-bit two's-complement
75 * numbers, -x is represented as
76 *
77 * x 32
78 * ( 2 - ------ ) mod 2 * 2
79 * 32
80 * 2
81 *
82 * (the `mod 2' subtracts 1 from 1.bbbb). To avoid lots of 2^32s,
83 * we can treat this as if the radix point were just to the left
84 * of the sign bit (multiply by 2^32), and get
85 *
86 * -x = (2 - x) mod 2
87 *
88 * Then, ignoring the `mod 2's for convenience:
89 *
90 * x * y = xy
91 * -x * y = 2y - xy
92 * x * -y = 2x - xy
93 * -x * -y = 4 - 2x - 2y + xy
94 *
95 * For signed multiplies, we subtract (x << 32) from the partial
96 * product to fix this problem for negative multipliers (see mul.s).
97 * Because of the way the shift into the partial product is calculated
98 * (N xor V), this term is automatically removed for the multiplicand,
99 * so we don't have to adjust.
100 *
101 * But for unsigned multiplies, the high order bit wasn't a sign bit,
102 * and the correction is wrong. So for unsigned multiplies where the
103 * high order bit is one, we end up with xy - (y << 32). To fix it
104 * we add y << 32.
105 */
106#if 0
107 tst %o1
108 bl,a 1f ! if %o1 < 0 (high order bit = 1),
109 add %o4, %o0, %o4 ! %o4 += %o0 (add y to upper half)
110
1111:
112 rd %y, %o0 ! get lower half of product
113 retl
114 addcc %o4, %g0, %o1 ! put upper half in place and set Z for %o1==0
115#else
116 /* Faster code from tege@sics.se. */
117 sra %o1, 31, %o2 ! make mask from sign bit
118 and %o0, %o2, %o2 ! %o2 = 0 or %o0, depending on sign of %o1
119 rd %y, %o0 ! get lower half of product
120 retl
121 addcc %o4, %o2, %o1 ! add compensation and put upper half in place
122#endif
123
124Lmul_shortway:
125 /*
126 * Short multiply. 12 steps, followed by a final shift step.
127 * The resulting bits are off by 12 and (32-12) = 20 bit positions,
128 * but there is no problem with %o0 being negative (unlike above),
129 * and overflow is impossible (the answer is at most 24 bits long).
130 */
131 mulscc %o4, %o1, %o4 ! 1
132 mulscc %o4, %o1, %o4 ! 2
133 mulscc %o4, %o1, %o4 ! 3
134 mulscc %o4, %o1, %o4 ! 4
135 mulscc %o4, %o1, %o4 ! 5
136 mulscc %o4, %o1, %o4 ! 6
137 mulscc %o4, %o1, %o4 ! 7
138 mulscc %o4, %o1, %o4 ! 8
139 mulscc %o4, %o1, %o4 ! 9
140 mulscc %o4, %o1, %o4 ! 10
141 mulscc %o4, %o1, %o4 ! 11
142 mulscc %o4, %o1, %o4 ! 12
143 mulscc %o4, %g0, %o4 ! final shift
144
145 /*
146 * %o4 has 20 of the bits that should be in the result; %y has
147 * the bottom 12 (as %y's top 12). That is:
148 *
149 * %o4 %y
150 * +----------------+----------------+
151 * | -12- | -20- | -12- | -20- |
152 * +------(---------+------)---------+
153 * -----result-----
154 *
155 * The 12 bits of %o4 left of the `result' area are all zero;
156 * in fact, all top 20 bits of %o4 are zero.
157 */
158
159 rd %y, %o5
160 sll %o4, 12, %o0 ! shift middle bits left 12
161 srl %o5, 20, %o5 ! shift low bits right 20
162 or %o5, %o0, %o0
163 retl
164 addcc %g0, %g0, %o1 ! %o1 = zero, and set Z
165
166 .globl .umul_patch
167.umul_patch:
168 umul %o0, %o1, %o0
169 retl
170 rd %y, %o1
171 nop