1 /* $Id: umul.S,v 1.4 1996/09/30 02:22:39 davem Exp $
2 * umul.S: This routine was taken from glibc-1.09 and is covered
3 * by the GNU Library General Public License Version 2.
8 * Unsigned multiply. Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the
9 * upper 32 bits of the 64-bit product).
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.
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:
20 * bnz overflow (or tnz)
26 _Umul: /* needed for export */
28 mov %o0, %y ! multiplier -> Y
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
35 * Long multiply. 32 steps, followed by a final shift step.
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
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
78 * ( 2 - ------ ) mod 2 * 2
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
88 * Then, ignoring the `mod 2's for convenience:
93 * -x * -y = 4 - 2x - 2y + xy
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.
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
108 bl,a 1f ! if %o1 < 0 (high order bit = 1),
109 add %o4, %o0, %o4 ! %o4 += %o0 (add y to upper half)
112 rd %y, %o0 ! get lower half of product
114 addcc %o4, %g0, %o1 ! put upper half in place and set Z for %o1==0
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
121 addcc %o4, %o2, %o1 ! add compensation and put upper half in place
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).
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
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:
150 * +----------------+----------------+
151 * | -12- | -20- | -12- | -20- |
152 * +------(---------+------)---------+
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.
160 sll %o4, 12, %o0 ! shift middle bits left 12
161 srl %o5, 20, %o5 ! shift low bits right 20
164 addcc %g0, %g0, %o1 ! %o1 = zero, and set Z