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 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 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