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5 * This software was developed by the Computer Systems Engineering group
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37 * @(#)qdivrem.c 8.1 (Berkeley) 6/4/93
38 * $DragonFly: src/lib/libc/quad/qdivrem.c,v 1.3 2004/10/25 19:38:01 drhodus Exp $
42 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
43 * section 4.3.1, pp. 257--259.
48 #define B (1 << HALF_BITS) /* digit base */
50 /* Combine two `digits' to make a single two-digit number. */
51 #define COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b))
53 /* select a type for digits in base B: use unsigned short if they fit */
54 #if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
55 typedef unsigned short digit;
61 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
62 * `fall out' the left (there never will be any such anyway).
63 * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS.
66 shl(digit *p, int len, int sh)
70 for (i = 0; i < len; i++)
71 p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
72 p[i] = LHALF(p[i] << sh);
76 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
78 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
79 * fit within u_long. As a consequence, the maximum length dividend and
80 * divisor are 4 `digits' in this base (they are shorter if they have
84 __qdivrem(uq, vq, arq)
85 u_quad_t uq, vq, *arq;
92 digit uspace[5], vspace[5], qspace[5];
95 * Take care of special cases: divide by zero, and u < v.
99 static volatile const unsigned int zero = 0;
101 tmp.ul[H] = tmp.ul[L] = 1 / zero;
116 * Break dividend and divisor into digits in base B, then
117 * count leading zeros to determine m and n. When done, we
119 * u = (u[1]u[2]...u[m+n]) sub B
120 * v = (v[1]v[2]...v[n]) sub B
122 * 1 < n <= 4 (if n = 1, we use a different division algorithm)
123 * m >= 0 (otherwise u < v, which we already checked)
130 u[1] = HHALF(tmp.ul[H]);
131 u[2] = LHALF(tmp.ul[H]);
132 u[3] = HHALF(tmp.ul[L]);
133 u[4] = LHALF(tmp.ul[L]);
135 v[1] = HHALF(tmp.ul[H]);
136 v[2] = LHALF(tmp.ul[H]);
137 v[3] = HHALF(tmp.ul[L]);
138 v[4] = LHALF(tmp.ul[L]);
139 for (n = 4; v[1] == 0; v++) {
141 u_long rbj; /* r*B+u[j] (not root boy jim) */
142 digit q1, q2, q3, q4;
145 * Change of plan, per exercise 16.
148 * q[j] = floor((r*B + u[j]) / v),
149 * r = (r*B + u[j]) % v;
150 * We unroll this completely here.
152 t = v[2]; /* nonzero, by definition */
154 rbj = COMBINE(u[1] % t, u[2]);
156 rbj = COMBINE(rbj % t, u[3]);
158 rbj = COMBINE(rbj % t, u[4]);
162 tmp.ul[H] = COMBINE(q1, q2);
163 tmp.ul[L] = COMBINE(q3, q4);
169 * By adjusting q once we determine m, we can guarantee that
170 * there is a complete four-digit quotient at &qspace[1] when
173 for (m = 4 - n; u[1] == 0; u++)
175 for (i = 4 - m; --i >= 0;)
180 * Here we run Program D, translated from MIX to C and acquiring
181 * a few minor changes.
183 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
186 for (t = v[1]; t < B / 2; t <<= 1)
189 shl(&u[0], m + n, d); /* u <<= d */
190 shl(&v[1], n - 1, d); /* v <<= d */
196 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
197 v2 = v[2]; /* for D3 */
202 * D3: Calculate qhat (\^q, in TeX notation).
203 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
204 * let rhat = (u[j]*B + u[j+1]) mod v[1].
205 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
206 * decrement qhat and increase rhat correspondingly.
207 * Note that if rhat >= B, v[2]*qhat < rhat*B.
209 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
210 uj1 = u[j + 1]; /* for D3 only */
211 uj2 = u[j + 2]; /* for D3 only */
217 u_long n = COMBINE(uj0, uj1);
221 while (v2 * qhat > COMBINE(rhat, uj2)) {
224 if ((rhat += v1) >= B)
228 * D4: Multiply and subtract.
229 * The variable `t' holds any borrows across the loop.
230 * We split this up so that we do not require v[0] = 0,
231 * and to eliminate a final special case.
233 for (t = 0, i = n; i > 0; i--) {
234 t = u[i + j] - v[i] * qhat - t;
236 t = (B - HHALF(t)) & (B - 1);
241 * D5: test remainder.
242 * There is a borrow if and only if HHALF(t) is nonzero;
243 * in that (rare) case, qhat was too large (by exactly 1).
244 * Fix it by adding v[1..n] to u[j..j+n].
248 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
249 t += u[i + j] + v[i];
253 u[j] = LHALF(u[j] + t);
256 } while (++j <= m); /* D7: loop on j. */
259 * If caller wants the remainder, we have to calculate it as
260 * u[m..m+n] >> d (this is at most n digits and thus fits in
261 * u[m+1..m+n], but we may need more source digits).
265 for (i = m + n; i > m; --i)
267 LHALF(u[i - 1] << (HALF_BITS - d));
270 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
271 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
275 tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
276 tmp.ul[L] = COMBINE(qspace[3], qspace[4]);