2 * Copyright (c) 1992, 1993
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5 * This software was developed by the Computer Systems Engineering group
6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
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37 * From: Id: qdivrem.c,v 1.7 1997/11/07 09:20:40 phk Exp
40 #include <sys/cdefs.h>
41 __FBSDID("$FreeBSD$");
44 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
45 * section 4.3.1, pp. 257--259.
50 #define B (1 << HALF_BITS) /* digit base */
52 /* Combine two `digits' to make a single two-digit number. */
53 #define COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b))
55 /* select a type for digits in base B: use unsigned short if they fit */
56 #if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
57 typedef unsigned short digit;
63 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
64 * `fall out' the left (there never will be any such anyway).
65 * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS.
68 shl(digit *p, int len, int sh)
72 for (i = 0; i < len; i++)
73 p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
74 p[i] = LHALF(p[i] << sh);
78 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
80 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
81 * fit within u_long. As a consequence, the maximum length dividend and
82 * divisor are 4 `digits' in this base (they are shorter if they have
86 __qdivrem(uq, vq, arq)
87 u_quad_t uq, vq, *arq;
94 digit uspace[5], vspace[5], qspace[5];
97 * Take care of special cases: divide by zero, and u < v.
100 /* divide by zero. */
101 static volatile const unsigned int zero = 0;
103 tmp.ul[H] = tmp.ul[L] = 1 / zero;
118 * Break dividend and divisor into digits in base B, then
119 * count leading zeros to determine m and n. When done, we
121 * u = (u[1]u[2]...u[m+n]) sub B
122 * v = (v[1]v[2]...v[n]) sub B
124 * 1 < n <= 4 (if n = 1, we use a different division algorithm)
125 * m >= 0 (otherwise u < v, which we already checked)
132 u[1] = HHALF(tmp.ul[H]);
133 u[2] = LHALF(tmp.ul[H]);
134 u[3] = HHALF(tmp.ul[L]);
135 u[4] = LHALF(tmp.ul[L]);
137 v[1] = HHALF(tmp.ul[H]);
138 v[2] = LHALF(tmp.ul[H]);
139 v[3] = HHALF(tmp.ul[L]);
140 v[4] = LHALF(tmp.ul[L]);
141 for (n = 4; v[1] == 0; v++) {
143 u_long rbj; /* r*B+u[j] (not root boy jim) */
144 digit q1, q2, q3, q4;
147 * Change of plan, per exercise 16.
150 * q[j] = floor((r*B + u[j]) / v),
151 * r = (r*B + u[j]) % v;
152 * We unroll this completely here.
154 t = v[2]; /* nonzero, by definition */
156 rbj = COMBINE(u[1] % t, u[2]);
158 rbj = COMBINE(rbj % t, u[3]);
160 rbj = COMBINE(rbj % t, u[4]);
164 tmp.ul[H] = COMBINE(q1, q2);
165 tmp.ul[L] = COMBINE(q3, q4);
171 * By adjusting q once we determine m, we can guarantee that
172 * there is a complete four-digit quotient at &qspace[1] when
175 for (m = 4 - n; u[1] == 0; u++)
177 for (i = 4 - m; --i >= 0;)
182 * Here we run Program D, translated from MIX to C and acquiring
183 * a few minor changes.
185 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
188 for (t = v[1]; t < B / 2; t <<= 1)
191 shl(&u[0], m + n, d); /* u <<= d */
192 shl(&v[1], n - 1, d); /* v <<= d */
198 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
199 v2 = v[2]; /* for D3 */
204 * D3: Calculate qhat (\^q, in TeX notation).
205 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
206 * let rhat = (u[j]*B + u[j+1]) mod v[1].
207 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
208 * decrement qhat and increase rhat correspondingly.
209 * Note that if rhat >= B, v[2]*qhat < rhat*B.
211 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
212 uj1 = u[j + 1]; /* for D3 only */
213 uj2 = u[j + 2]; /* for D3 only */
219 u_long nn = COMBINE(uj0, uj1);
223 while (v2 * qhat > COMBINE(rhat, uj2)) {
226 if ((rhat += v1) >= B)
230 * D4: Multiply and subtract.
231 * The variable `t' holds any borrows across the loop.
232 * We split this up so that we do not require v[0] = 0,
233 * and to eliminate a final special case.
235 for (t = 0, i = n; i > 0; i--) {
236 t = u[i + j] - v[i] * qhat - t;
238 t = (B - HHALF(t)) & (B - 1);
243 * D5: test remainder.
244 * There is a borrow if and only if HHALF(t) is nonzero;
245 * in that (rare) case, qhat was too large (by exactly 1).
246 * Fix it by adding v[1..n] to u[j..j+n].
250 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
251 t += u[i + j] + v[i];
255 u[j] = LHALF(u[j] + t);
258 } while (++j <= m); /* D7: loop on j. */
261 * If caller wants the remainder, we have to calculate it as
262 * u[m..m+n] >> d (this is at most n digits and thus fits in
263 * u[m+1..m+n], but we may need more source digits).
267 for (i = m + n; i > m; --i)
269 LHALF(u[i - 1] << (HALF_BITS - d));
272 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
273 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
277 tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
278 tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
283 * Divide two unsigned quads.
291 return (__qdivrem(a, b, (u_quad_t *)0));
295 * Return remainder after dividing two unsigned quads.
303 (void)__qdivrem(a, b, &r);
308 * Divide two signed quads.
309 * ??? if -1/2 should produce -1 on this machine, this code is wrong
319 ua = -(u_quad_t)a, neg = 1;
323 ub = -(u_quad_t)b, neg ^= 1;
326 uq = __qdivrem(ua, ub, (u_quad_t *)0);
327 return (neg ? -uq : uq);
331 * Return remainder after dividing two signed quads.
334 * If -1/2 should produce -1 on this machine, this code is wrong.
344 ua = -(u_quad_t)a, neg = 1;
351 (void)__qdivrem(ua, ub, &ur);
352 return (neg ? -ur : ur);