Home | History | Annotate | Line # | Download | only in quad
      1 /*	$NetBSD: qdivrem.c,v 1.4 2012/03/20 16:21:41 matt Exp $	*/
      2 
      3 /*-
      4  * Copyright (c) 1992, 1993
      5  *	The Regents of the University of California.  All rights reserved.
      6  *
      7  * This software was developed by the Computer Systems Engineering group
      8  * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
      9  * contributed to Berkeley.
     10  *
     11  * Redistribution and use in source and binary forms, with or without
     12  * modification, are permitted provided that the following conditions
     13  * are met:
     14  * 1. Redistributions of source code must retain the above copyright
     15  *    notice, this list of conditions and the following disclaimer.
     16  * 2. Redistributions in binary form must reproduce the above copyright
     17  *    notice, this list of conditions and the following disclaimer in the
     18  *    documentation and/or other materials provided with the distribution.
     19  * 3. Neither the name of the University nor the names of its contributors
     20  *    may be used to endorse or promote products derived from this software
     21  *    without specific prior written permission.
     22  *
     23  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
     24  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
     25  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
     26  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
     27  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
     28  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
     29  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
     30  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
     31  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
     32  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
     33  * SUCH DAMAGE.
     34  */
     35 
     36 #include <sys/cdefs.h>
     37 #if defined(LIBC_SCCS) && !defined(lint)
     38 #if 0
     39 static char sccsid[] = "@(#)qdivrem.c	8.1 (Berkeley) 6/4/93";
     40 #else
     41 __RCSID("$NetBSD: qdivrem.c,v 1.4 2012/03/20 16:21:41 matt Exp $");
     42 #endif
     43 #endif /* LIBC_SCCS and not lint */
     44 
     45 /*
     46  * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
     47  * section 4.3.1, pp. 257--259.
     48  */
     49 
     50 #include "quad.h"
     51 
     52 #define	B	((int)1 << (unsigned int)HALF_BITS)	/* digit base */
     53 
     54 /* Combine two `digits' to make a single two-digit number. */
     55 #define	COMBINE(a, b) (((u_int)(a) << (unsigned int)HALF_BITS) | (b))
     56 
     57 /* select a type for digits in base B: use unsigned short if they fit */
     58 #if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
     59 typedef unsigned short digit;
     60 #else
     61 typedef u_int digit;
     62 #endif
     63 
     64 static void shl(digit *p, int len, int sh);
     65 
     66 /*
     67  * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
     68  *
     69  * We do this in base 2-sup-HALF_BITS, so that all intermediate products
     70  * fit within u_int.  As a consequence, the maximum length dividend and
     71  * divisor are 4 `digits' in this base (they are shorter if they have
     72  * leading zeros).
     73  */
     74 u_quad_t
     75 __qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
     76 {
     77 	union uu tmp;
     78 	digit *u, *v, *q;
     79 	digit v1, v2;
     80 	u_int qhat, rhat, t;
     81 	int m, n, d, j, i;
     82 	digit uspace[5], vspace[5], qspace[5];
     83 
     84 	/*
     85 	 * Take care of special cases: divide by zero, and u < v.
     86 	 */
     87 	if (vq == 0) {
     88 		/* divide by zero. */
     89 		static volatile const unsigned int zero = 0;
     90 
     91 		tmp.ul[H] = tmp.ul[L] = 1 / zero;
     92 		if (arq)
     93 			*arq = uq;
     94 		return (tmp.q);
     95 	}
     96 	if (uq < vq) {
     97 		if (arq)
     98 			*arq = uq;
     99 		return (0);
    100 	}
    101 	u = &uspace[0];
    102 	v = &vspace[0];
    103 	q = &qspace[0];
    104 
    105 	/*
    106 	 * Break dividend and divisor into digits in base B, then
    107 	 * count leading zeros to determine m and n.  When done, we
    108 	 * will have:
    109 	 *	u = (u[1]u[2]...u[m+n]) sub B
    110 	 *	v = (v[1]v[2]...v[n]) sub B
    111 	 *	v[1] != 0
    112 	 *	1 < n <= 4 (if n = 1, we use a different division algorithm)
    113 	 *	m >= 0 (otherwise u < v, which we already checked)
    114 	 *	m + n = 4
    115 	 * and thus
    116 	 *	m = 4 - n <= 2
    117 	 */
    118 	tmp.uq = uq;
    119 	u[0] = 0;
    120 	u[1] = (digit)HHALF(tmp.ul[H]);
    121 	u[2] = (digit)LHALF(tmp.ul[H]);
    122 	u[3] = (digit)HHALF(tmp.ul[L]);
    123 	u[4] = (digit)LHALF(tmp.ul[L]);
    124 	tmp.uq = vq;
    125 	v[1] = (digit)HHALF(tmp.ul[H]);
    126 	v[2] = (digit)LHALF(tmp.ul[H]);
    127 	v[3] = (digit)HHALF(tmp.ul[L]);
    128 	v[4] = (digit)LHALF(tmp.ul[L]);
    129 	for (n = 4; v[1] == 0; v++) {
    130 		if (--n == 1) {
    131 			u_int rbj;	/* r*B+u[j] (not root boy jim) */
    132 			digit q1, q2, q3, q4;
    133 
    134 			/*
    135 			 * Change of plan, per exercise 16.
    136 			 *	r = 0;
    137 			 *	for j = 1..4:
    138 			 *		q[j] = floor((r*B + u[j]) / v),
    139 			 *		r = (r*B + u[j]) % v;
    140 			 * We unroll this completely here.
    141 			 */
    142 			t = v[2];	/* nonzero, by definition */
    143 			q1 = (digit)(u[1] / t);
    144 			rbj = COMBINE(u[1] % t, u[2]);
    145 			q2 = (digit)(rbj / t);
    146 			rbj = COMBINE(rbj % t, u[3]);
    147 			q3 = (digit)(rbj / t);
    148 			rbj = COMBINE(rbj % t, u[4]);
    149 			q4 = (digit)(rbj / t);
    150 			if (arq)
    151 				*arq = rbj % t;
    152 			tmp.ul[H] = COMBINE(q1, q2);
    153 			tmp.ul[L] = COMBINE(q3, q4);
    154 			return (tmp.q);
    155 		}
    156 	}
    157 
    158 	/*
    159 	 * By adjusting q once we determine m, we can guarantee that
    160 	 * there is a complete four-digit quotient at &qspace[1] when
    161 	 * we finally stop.
    162 	 */
    163 	for (m = 4 - n; u[1] == 0; u++)
    164 		m--;
    165 	for (i = 4 - m; --i >= 0;)
    166 		q[i] = 0;
    167 	q += 4 - m;
    168 
    169 	/*
    170 	 * Here we run Program D, translated from MIX to C and acquiring
    171 	 * a few minor changes.
    172 	 *
    173 	 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
    174 	 */
    175 	d = 0;
    176 	for (t = v[1]; t < B / 2; t <<= (unsigned int)1)
    177 		d++;
    178 	if (d > 0) {
    179 		shl(&u[0], m + n, d);		/* u <<= d */
    180 		shl(&v[1], n - 1, d);		/* v <<= d */
    181 	}
    182 	/*
    183 	 * D2: j = 0.
    184 	 */
    185 	j = 0;
    186 	v1 = v[1];	/* for D3 -- note that v[1..n] are constant */
    187 	v2 = v[2];	/* for D3 */
    188 	do {
    189 		digit uj0, uj1, uj2;
    190 
    191 		/*
    192 		 * D3: Calculate qhat (\^q, in TeX notation).
    193 		 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
    194 		 * let rhat = (u[j]*B + u[j+1]) mod v[1].
    195 		 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
    196 		 * decrement qhat and increase rhat correspondingly.
    197 		 * Note that if rhat >= B, v[2]*qhat < rhat*B.
    198 		 */
    199 		uj0 = u[j + 0];	/* for D3 only -- note that u[j+...] change */
    200 		uj1 = u[j + 1];	/* for D3 only */
    201 		uj2 = u[j + 2];	/* for D3 only */
    202 		if (uj0 == v1) {
    203 			qhat = B;
    204 			rhat = uj1;
    205 			goto qhat_too_big;
    206 		} else {
    207 			u_int nn = COMBINE(uj0, uj1);
    208 			qhat = nn / v1;
    209 			rhat = nn % v1;
    210 		}
    211 		while (v2 * qhat > COMBINE(rhat, uj2)) {
    212 	qhat_too_big:
    213 			qhat--;
    214 			if ((rhat += v1) >= B)
    215 				break;
    216 		}
    217 		/*
    218 		 * D4: Multiply and subtract.
    219 		 * The variable `t' holds any borrows across the loop.
    220 		 * We split this up so that we do not require v[0] = 0,
    221 		 * and to eliminate a final special case.
    222 		 */
    223 		for (t = 0, i = n; i > 0; i--) {
    224 			t = u[i + j] - v[i] * qhat - t;
    225 			u[i + j] = (digit)LHALF(t);
    226 			t = (B - HHALF(t)) & (B - 1);
    227 		}
    228 		t = u[j] - t;
    229 		u[j] = (digit)LHALF(t);
    230 		/*
    231 		 * D5: test remainder.
    232 		 * There is a borrow if and only if HHALF(t) is nonzero;
    233 		 * in that (rare) case, qhat was too large (by exactly 1).
    234 		 * Fix it by adding v[1..n] to u[j..j+n].
    235 		 */
    236 		if (HHALF(t)) {
    237 			qhat--;
    238 			for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
    239 				t += u[i + j] + v[i];
    240 				u[i + j] = (digit)LHALF(t);
    241 				t = HHALF(t);
    242 			}
    243 			u[j] = (digit)LHALF(u[j] + t);
    244 		}
    245 		q[j] = (digit)qhat;
    246 	} while (++j <= m);		/* D7: loop on j. */
    247 
    248 	/*
    249 	 * If caller wants the remainder, we have to calculate it as
    250 	 * u[m..m+n] >> d (this is at most n digits and thus fits in
    251 	 * u[m+1..m+n], but we may need more source digits).
    252 	 */
    253 	if (arq) {
    254 		if (d) {
    255 			for (i = m + n; i > m; --i)
    256 				u[i] = (digit)(((u_int)u[i] >> d) |
    257 				    LHALF((u_int)u[i - 1] << (unsigned int)(HALF_BITS - d)));
    258 			u[i] = 0;
    259 		}
    260 		tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
    261 		tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
    262 		*arq = tmp.q;
    263 	}
    264 
    265 	tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
    266 	tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
    267 	return (tmp.q);
    268 }
    269 
    270 /*
    271  * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
    272  * `fall out' the left (there never will be any such anyway).
    273  * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
    274  */
    275 static void
    276 shl(digit *p, int len, int sh)
    277 {
    278 	int i;
    279 
    280 	for (i = 0; i < len; i++)
    281 		p[i] = (digit)(LHALF((u_int)p[i] << sh) |
    282 		    ((u_int)p[i + 1] >> (HALF_BITS - sh)));
    283 	p[i] = (digit)(LHALF((u_int)p[i] << sh));
    284 }
    285