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 mov %o0, %y ! multiplier -> Y
28 andncc %o4, 0xfff, %g0 ! test bits 12..31 of *both* args
29 be Lmul_shortway ! if zero, can do it the short way
30 andcc %g0, %g0, %o4 ! zero the partial product and clear N and V
33 * Long multiply. 32 steps, followed by a final shift step.
35 mulscc %o4, %o1, %o4 ! 1
36 mulscc %o4, %o1, %o4 ! 2
37 mulscc %o4, %o1, %o4 ! 3
38 mulscc %o4, %o1, %o4 ! 4
39 mulscc %o4, %o1, %o4 ! 5
40 mulscc %o4, %o1, %o4 ! 6
41 mulscc %o4, %o1, %o4 ! 7
42 mulscc %o4, %o1, %o4 ! 8
43 mulscc %o4, %o1, %o4 ! 9
44 mulscc %o4, %o1, %o4 ! 10
45 mulscc %o4, %o1, %o4 ! 11
46 mulscc %o4, %o1, %o4 ! 12
47 mulscc %o4, %o1, %o4 ! 13
48 mulscc %o4, %o1, %o4 ! 14
49 mulscc %o4, %o1, %o4 ! 15
50 mulscc %o4, %o1, %o4 ! 16
51 mulscc %o4, %o1, %o4 ! 17
52 mulscc %o4, %o1, %o4 ! 18
53 mulscc %o4, %o1, %o4 ! 19
54 mulscc %o4, %o1, %o4 ! 20
55 mulscc %o4, %o1, %o4 ! 21
56 mulscc %o4, %o1, %o4 ! 22
57 mulscc %o4, %o1, %o4 ! 23
58 mulscc %o4, %o1, %o4 ! 24
59 mulscc %o4, %o1, %o4 ! 25
60 mulscc %o4, %o1, %o4 ! 26
61 mulscc %o4, %o1, %o4 ! 27
62 mulscc %o4, %o1, %o4 ! 28
63 mulscc %o4, %o1, %o4 ! 29
64 mulscc %o4, %o1, %o4 ! 30
65 mulscc %o4, %o1, %o4 ! 31
66 mulscc %o4, %o1, %o4 ! 32
67 mulscc %o4, %g0, %o4 ! final shift
71 * Normally, with the shift-and-add approach, if both numbers are
72 * positive you get the correct result. With 32-bit two's-complement
73 * numbers, -x is represented as
76 * ( 2 - ------ ) mod 2 * 2
80 * (the `mod 2' subtracts 1 from 1.bbbb). To avoid lots of 2^32s,
81 * we can treat this as if the radix point were just to the left
82 * of the sign bit (multiply by 2^32), and get
86 * Then, ignoring the `mod 2's for convenience:
91 * -x * -y = 4 - 2x - 2y + xy
93 * For signed multiplies, we subtract (x << 32) from the partial
94 * product to fix this problem for negative multipliers (see mul.s).
95 * Because of the way the shift into the partial product is calculated
96 * (N xor V), this term is automatically removed for the multiplicand,
97 * so we don't have to adjust.
99 * But for unsigned multiplies, the high order bit wasn't a sign bit,
100 * and the correction is wrong. So for unsigned multiplies where the
101 * high order bit is one, we end up with xy - (y << 32). To fix it
106 bl,a 1f ! if %o1 < 0 (high order bit = 1),
107 add %o4, %o0, %o4 ! %o4 += %o0 (add y to upper half)
110 rd %y, %o0 ! get lower half of product
112 addcc %o4, %g0, %o1 ! put upper half in place and set Z for %o1==0
114 /* Faster code from tege@sics.se. */
115 sra %o1, 31, %o2 ! make mask from sign bit
116 and %o0, %o2, %o2 ! %o2 = 0 or %o0, depending on sign of %o1
117 rd %y, %o0 ! get lower half of product
119 addcc %o4, %o2, %o1 ! add compensation and put upper half in place
124 * Short multiply. 12 steps, followed by a final shift step.
125 * The resulting bits are off by 12 and (32-12) = 20 bit positions,
126 * but there is no problem with %o0 being negative (unlike above),
127 * and overflow is impossible (the answer is at most 24 bits long).
129 mulscc %o4, %o1, %o4 ! 1
130 mulscc %o4, %o1, %o4 ! 2
131 mulscc %o4, %o1, %o4 ! 3
132 mulscc %o4, %o1, %o4 ! 4
133 mulscc %o4, %o1, %o4 ! 5
134 mulscc %o4, %o1, %o4 ! 6
135 mulscc %o4, %o1, %o4 ! 7
136 mulscc %o4, %o1, %o4 ! 8
137 mulscc %o4, %o1, %o4 ! 9
138 mulscc %o4, %o1, %o4 ! 10
139 mulscc %o4, %o1, %o4 ! 11
140 mulscc %o4, %o1, %o4 ! 12
141 mulscc %o4, %g0, %o4 ! final shift
144 * %o4 has 20 of the bits that should be in the result; %y has
145 * the bottom 12 (as %y's top 12). That is:
148 * +----------------+----------------+
149 * | -12- | -20- | -12- | -20- |
150 * +------(---------+------)---------+
153 * The 12 bits of %o4 left of the `result' area are all zero;
154 * in fact, all top 20 bits of %o4 are zero.
158 sll %o4, 12, %o0 ! shift middle bits left 12
159 srl %o5, 20, %o5 ! shift low bits right 20
162 addcc %g0, %g0, %o1 ! %o1 = zero, and set Z