diff options
author | Jonathan Herman <hermanjl@cs.unc.edu> | 2013-01-22 10:38:37 -0500 |
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committer | Jonathan Herman <hermanjl@cs.unc.edu> | 2013-01-22 10:38:37 -0500 |
commit | fcc9d2e5a6c89d22b8b773a64fb4ad21ac318446 (patch) | |
tree | a57612d1888735a2ec7972891b68c1ac5ec8faea /arch/sparc/lib/umul.S | |
parent | 8dea78da5cee153b8af9c07a2745f6c55057fe12 (diff) |
Diffstat (limited to 'arch/sparc/lib/umul.S')
-rw-r--r-- | arch/sparc/lib/umul.S | 171 |
1 files changed, 171 insertions, 0 deletions
diff --git a/arch/sparc/lib/umul.S b/arch/sparc/lib/umul.S new file mode 100644 index 00000000000..1f36ae68252 --- /dev/null +++ b/arch/sparc/lib/umul.S | |||
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1 | /* | ||
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 | ||
24 | .globl _Umul | ||
25 | .umul: | ||
26 | _Umul: /* needed for export */ | ||
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 | |||
111 | 1: | ||
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 | |||
124 | Lmul_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 | ||