1 /*-
2 * Copyright (c) 1992, 1993
3 * The Regents of the University of California. All rights reserved.
4 *
5 * This software was developed by the Computer Systems Engineering group
6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
7 * contributed to Berkeley.
8 *
9 * %sccs.include.redist.c%
10 */
11
12 #if defined(LIBC_SCCS) && !defined(lint)
13 static char sccsid[] = "@(#)muldi3.c 8.1 (Berkeley) 06/04/93";
14 #endif /* LIBC_SCCS and not lint */
15
16 #include "quad.h"
17
18 /*
19 * Multiply two quads.
20 *
21 * Our algorithm is based on the following. Split incoming quad values
22 * u and v (where u,v >= 0) into
23 *
24 * u = 2^n u1 * u0 (n = number of bits in `u_long', usu. 32)
25 *
26 * and
27 *
28 * v = 2^n v1 * v0
29 *
30 * Then
31 *
32 * uv = 2^2n u1 v1 + 2^n u1 v0 + 2^n v1 u0 + u0 v0
33 * = 2^2n u1 v1 + 2^n (u1 v0 + v1 u0) + u0 v0
34 *
35 * Now add 2^n u1 v1 to the first term and subtract it from the middle,
36 * and add 2^n u0 v0 to the last term and subtract it from the middle.
37 * This gives:
38 *
39 * uv = (2^2n + 2^n) (u1 v1) +
40 * (2^n) (u1 v0 - u1 v1 + u0 v1 - u0 v0) +
41 * (2^n + 1) (u0 v0)
42 *
43 * Factoring the middle a bit gives us:
44 *
45 * uv = (2^2n + 2^n) (u1 v1) + [u1v1 = high]
46 * (2^n) (u1 - u0) (v0 - v1) + [(u1-u0)... = mid]
47 * (2^n + 1) (u0 v0) [u0v0 = low]
48 *
49 * The terms (u1 v1), (u1 - u0) (v0 - v1), and (u0 v0) can all be done
50 * in just half the precision of the original. (Note that either or both
51 * of (u1 - u0) or (v0 - v1) may be negative.)
52 *
53 * This algorithm is from Knuth vol. 2 (2nd ed), section 4.3.3, p. 278.
54 *
55 * Since C does not give us a `long * long = quad' operator, we split
56 * our input quads into two longs, then split the two longs into two
57 * shorts. We can then calculate `short * short = long' in native
58 * arithmetic.
59 *
60 * Our product should, strictly speaking, be a `long quad', with 128
61 * bits, but we are going to discard the upper 64. In other words,
62 * we are not interested in uv, but rather in (uv mod 2^2n). This
63 * makes some of the terms above vanish, and we get:
64 *
65 * (2^n)(high) + (2^n)(mid) + (2^n + 1)(low)
66 *
67 * or
68 *
69 * (2^n)(high + mid + low) + low
70 *
71 * Furthermore, `high' and `mid' can be computed mod 2^n, as any factor
72 * of 2^n in either one will also vanish. Only `low' need be computed
73 * mod 2^2n, and only because of the final term above.
74 */
75 static quad_t __lmulq(u_long, u_long);
76
77 quad_t
__muldi3(a,b)78 __muldi3(a, b)
79 quad_t a, b;
80 {
81 union uu u, v, low, prod;
82 register u_long high, mid, udiff, vdiff;
83 register int negall, negmid;
84 #define u1 u.ul[H]
85 #define u0 u.ul[L]
86 #define v1 v.ul[H]
87 #define v0 v.ul[L]
88
89 /*
90 * Get u and v such that u, v >= 0. When this is finished,
91 * u1, u0, v1, and v0 will be directly accessible through the
92 * longword fields.
93 */
94 if (a >= 0)
95 u.q = a, negall = 0;
96 else
97 u.q = -a, negall = 1;
98 if (b >= 0)
99 v.q = b;
100 else
101 v.q = -b, negall ^= 1;
102
103 if (u1 == 0 && v1 == 0) {
104 /*
105 * An (I hope) important optimization occurs when u1 and v1
106 * are both 0. This should be common since most numbers
107 * are small. Here the product is just u0*v0.
108 */
109 prod.q = __lmulq(u0, v0);
110 } else {
111 /*
112 * Compute the three intermediate products, remembering
113 * whether the middle term is negative. We can discard
114 * any upper bits in high and mid, so we can use native
115 * u_long * u_long => u_long arithmetic.
116 */
117 low.q = __lmulq(u0, v0);
118
119 if (u1 >= u0)
120 negmid = 0, udiff = u1 - u0;
121 else
122 negmid = 1, udiff = u0 - u1;
123 if (v0 >= v1)
124 vdiff = v0 - v1;
125 else
126 vdiff = v1 - v0, negmid ^= 1;
127 mid = udiff * vdiff;
128
129 high = u1 * v1;
130
131 /*
132 * Assemble the final product.
133 */
134 prod.ul[H] = high + (negmid ? -mid : mid) + low.ul[L] +
135 low.ul[H];
136 prod.ul[L] = low.ul[L];
137 }
138 return (negall ? -prod.q : prod.q);
139 #undef u1
140 #undef u0
141 #undef v1
142 #undef v0
143 }
144
145 /*
146 * Multiply two 2N-bit longs to produce a 4N-bit quad, where N is half
147 * the number of bits in a long (whatever that is---the code below
148 * does not care as long as quad.h does its part of the bargain---but
149 * typically N==16).
150 *
151 * We use the same algorithm from Knuth, but this time the modulo refinement
152 * does not apply. On the other hand, since N is half the size of a long,
153 * we can get away with native multiplication---none of our input terms
154 * exceeds (ULONG_MAX >> 1).
155 *
156 * Note that, for u_long l, the quad-precision result
157 *
158 * l << N
159 *
160 * splits into high and low longs as HHALF(l) and LHUP(l) respectively.
161 */
162 static quad_t
__lmulq(u_long u,u_long v)163 __lmulq(u_long u, u_long v)
164 {
165 u_long u1, u0, v1, v0, udiff, vdiff, high, mid, low;
166 u_long prodh, prodl, was;
167 union uu prod;
168 int neg;
169
170 u1 = HHALF(u);
171 u0 = LHALF(u);
172 v1 = HHALF(v);
173 v0 = LHALF(v);
174
175 low = u0 * v0;
176
177 /* This is the same small-number optimization as before. */
178 if (u1 == 0 && v1 == 0)
179 return (low);
180
181 if (u1 >= u0)
182 udiff = u1 - u0, neg = 0;
183 else
184 udiff = u0 - u1, neg = 1;
185 if (v0 >= v1)
186 vdiff = v0 - v1;
187 else
188 vdiff = v1 - v0, neg ^= 1;
189 mid = udiff * vdiff;
190
191 high = u1 * v1;
192
193 /* prod = (high << 2N) + (high << N); */
194 prodh = high + HHALF(high);
195 prodl = LHUP(high);
196
197 /* if (neg) prod -= mid << N; else prod += mid << N; */
198 if (neg) {
199 was = prodl;
200 prodl -= LHUP(mid);
201 prodh -= HHALF(mid) + (prodl > was);
202 } else {
203 was = prodl;
204 prodl += LHUP(mid);
205 prodh += HHALF(mid) + (prodl < was);
206 }
207
208 /* prod += low << N */
209 was = prodl;
210 prodl += LHUP(low);
211 prodh += HHALF(low) + (prodl < was);
212 /* ... + low; */
213 if ((prodl += low) < low)
214 prodh++;
215
216 /* return 4N-bit product */
217 prod.ul[H] = prodh;
218 prod.ul[L] = prodl;
219 return (prod.q);
220 }
221