File: | src/sys/arch/amd64/stand/efiboot/bootx64/../../../../../lib/libkern/qdivrem.c |
Warning: | line 80, column 29 Division by zero |
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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 | * Redistribution and use in source and binary forms, with or without | |||
10 | * modification, are permitted provided that the following conditions | |||
11 | * are met: | |||
12 | * 1. Redistributions of source code must retain the above copyright | |||
13 | * notice, this list of conditions and the following disclaimer. | |||
14 | * 2. Redistributions in binary form must reproduce the above copyright | |||
15 | * notice, this list of conditions and the following disclaimer in the | |||
16 | * documentation and/or other materials provided with the distribution. | |||
17 | * 3. Neither the name of the University nor the names of its contributors | |||
18 | * may be used to endorse or promote products derived from this software | |||
19 | * without specific prior written permission. | |||
20 | * | |||
21 | * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND | |||
22 | * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE | |||
23 | * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE | |||
24 | * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE | |||
25 | * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL | |||
26 | * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS | |||
27 | * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) | |||
28 | * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT | |||
29 | * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY | |||
30 | * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF | |||
31 | * SUCH DAMAGE. | |||
32 | */ | |||
33 | ||||
34 | /* | |||
35 | * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed), | |||
36 | * section 4.3.1, pp. 257--259. | |||
37 | */ | |||
38 | ||||
39 | #include "quad.h" | |||
40 | ||||
41 | #define B((int)1 << (sizeof(int) * 8 / 2)) ((int)1 << HALF_BITS(sizeof(int) * 8 / 2)) /* digit base */ | |||
42 | ||||
43 | /* Combine two `digits' to make a single two-digit number. */ | |||
44 | #define COMBINE(a, b)(((u_int)(a) << (sizeof(int) * 8 / 2)) | (b)) (((u_int)(a) << HALF_BITS(sizeof(int) * 8 / 2)) | (b)) | |||
45 | ||||
46 | /* select a type for digits in base B: use unsigned short if they fit */ | |||
47 | #if UINT_MAX0xffffffffU == 0xffffffffU && USHRT_MAX0xffff >= 0xffff | |||
48 | typedef unsigned short digit; | |||
49 | #else | |||
50 | typedef u_int digit; | |||
51 | #endif | |||
52 | ||||
53 | static void shl(digit *p, int len, int sh); | |||
54 | ||||
55 | /* | |||
56 | * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v. | |||
57 | * | |||
58 | * We do this in base 2-sup-HALF_BITS, so that all intermediate products | |||
59 | * fit within u_int. As a consequence, the maximum length dividend and | |||
60 | * divisor are 4 `digits' in this base (they are shorter if they have | |||
61 | * leading zeros). | |||
62 | */ | |||
63 | u_quad_t | |||
64 | __qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq) | |||
65 | { | |||
66 | union uu tmp; | |||
67 | digit *u, *v, *q; | |||
68 | digit v1, v2; | |||
69 | u_int qhat, rhat, t; | |||
70 | int m, n, d, j, i; | |||
71 | digit uspace[5], vspace[5], qspace[5]; | |||
72 | ||||
73 | /* | |||
74 | * Take care of special cases: divide by zero, and u < v. | |||
75 | */ | |||
76 | if (vq == 0) { | |||
| ||||
77 | /* divide by zero. */ | |||
78 | static volatile const unsigned int zero = 0; | |||
79 | ||||
80 | tmp.ul[H1] = tmp.ul[L0] = 1 / zero; | |||
| ||||
81 | if (arq) | |||
82 | *arq = uq; | |||
83 | return (tmp.q); | |||
84 | } | |||
85 | if (uq < vq) { | |||
86 | if (arq) | |||
87 | *arq = uq; | |||
88 | return (0); | |||
89 | } | |||
90 | u = &uspace[0]; | |||
91 | v = &vspace[0]; | |||
92 | q = &qspace[0]; | |||
93 | ||||
94 | /* | |||
95 | * Break dividend and divisor into digits in base B, then | |||
96 | * count leading zeros to determine m and n. When done, we | |||
97 | * will have: | |||
98 | * u = (u[1]u[2]...u[m+n]) sub B | |||
99 | * v = (v[1]v[2]...v[n]) sub B | |||
100 | * v[1] != 0 | |||
101 | * 1 < n <= 4 (if n = 1, we use a different division algorithm) | |||
102 | * m >= 0 (otherwise u < v, which we already checked) | |||
103 | * m + n = 4 | |||
104 | * and thus | |||
105 | * m = 4 - n <= 2 | |||
106 | */ | |||
107 | tmp.uq = uq; | |||
108 | u[0] = 0; | |||
109 | u[1] = (digit)HHALF(tmp.ul[H])((u_int)(tmp.ul[1]) >> (sizeof(int) * 8 / 2)); | |||
110 | u[2] = (digit)LHALF(tmp.ul[H])((u_int)(tmp.ul[1]) & (((int)1 << (sizeof(int) * 8 / 2)) - 1)); | |||
111 | u[3] = (digit)HHALF(tmp.ul[L])((u_int)(tmp.ul[0]) >> (sizeof(int) * 8 / 2)); | |||
112 | u[4] = (digit)LHALF(tmp.ul[L])((u_int)(tmp.ul[0]) & (((int)1 << (sizeof(int) * 8 / 2)) - 1)); | |||
113 | tmp.uq = vq; | |||
114 | v[1] = (digit)HHALF(tmp.ul[H])((u_int)(tmp.ul[1]) >> (sizeof(int) * 8 / 2)); | |||
115 | v[2] = (digit)LHALF(tmp.ul[H])((u_int)(tmp.ul[1]) & (((int)1 << (sizeof(int) * 8 / 2)) - 1)); | |||
116 | v[3] = (digit)HHALF(tmp.ul[L])((u_int)(tmp.ul[0]) >> (sizeof(int) * 8 / 2)); | |||
117 | v[4] = (digit)LHALF(tmp.ul[L])((u_int)(tmp.ul[0]) & (((int)1 << (sizeof(int) * 8 / 2)) - 1)); | |||
118 | for (n = 4; v[1] == 0; v++) { | |||
119 | if (--n == 1) { | |||
120 | u_int rbj; /* r*B+u[j] (not root boy jim) */ | |||
121 | digit q1, q2, q3, q4; | |||
122 | ||||
123 | /* | |||
124 | * Change of plan, per exercise 16. | |||
125 | * r = 0; | |||
126 | * for j = 1..4: | |||
127 | * q[j] = floor((r*B + u[j]) / v), | |||
128 | * r = (r*B + u[j]) % v; | |||
129 | * We unroll this completely here. | |||
130 | */ | |||
131 | t = v[2]; /* nonzero, by definition */ | |||
132 | q1 = (digit)(u[1] / t); | |||
133 | rbj = COMBINE(u[1] % t, u[2])(((u_int)(u[1] % t) << (sizeof(int) * 8 / 2)) | (u[2])); | |||
134 | q2 = (digit)(rbj / t); | |||
135 | rbj = COMBINE(rbj % t, u[3])(((u_int)(rbj % t) << (sizeof(int) * 8 / 2)) | (u[3])); | |||
136 | q3 = (digit)(rbj / t); | |||
137 | rbj = COMBINE(rbj % t, u[4])(((u_int)(rbj % t) << (sizeof(int) * 8 / 2)) | (u[4])); | |||
138 | q4 = (digit)(rbj / t); | |||
139 | if (arq) | |||
140 | *arq = rbj % t; | |||
141 | tmp.ul[H1] = COMBINE(q1, q2)(((u_int)(q1) << (sizeof(int) * 8 / 2)) | (q2)); | |||
142 | tmp.ul[L0] = COMBINE(q3, q4)(((u_int)(q3) << (sizeof(int) * 8 / 2)) | (q4)); | |||
143 | return (tmp.q); | |||
144 | } | |||
145 | } | |||
146 | ||||
147 | /* | |||
148 | * By adjusting q once we determine m, we can guarantee that | |||
149 | * there is a complete four-digit quotient at &qspace[1] when | |||
150 | * we finally stop. | |||
151 | */ | |||
152 | for (m = 4 - n; u[1] == 0; u++) | |||
153 | m--; | |||
154 | for (i = 4 - m; --i >= 0;) | |||
155 | q[i] = 0; | |||
156 | q += 4 - m; | |||
157 | ||||
158 | /* | |||
159 | * Here we run Program D, translated from MIX to C and acquiring | |||
160 | * a few minor changes. | |||
161 | * | |||
162 | * D1: choose multiplier 1 << d to ensure v[1] >= B/2. | |||
163 | */ | |||
164 | d = 0; | |||
165 | for (t = v[1]; t < B((int)1 << (sizeof(int) * 8 / 2)) / 2; t <<= 1) | |||
166 | d++; | |||
167 | if (d > 0) { | |||
168 | shl(&u[0], m + n, d); /* u <<= d */ | |||
169 | shl(&v[1], n - 1, d); /* v <<= d */ | |||
170 | } | |||
171 | /* | |||
172 | * D2: j = 0. | |||
173 | */ | |||
174 | j = 0; | |||
175 | v1 = v[1]; /* for D3 -- note that v[1..n] are constant */ | |||
176 | v2 = v[2]; /* for D3 */ | |||
177 | do { | |||
178 | digit uj0, uj1, uj2; | |||
179 | ||||
180 | /* | |||
181 | * D3: Calculate qhat (\^q, in TeX notation). | |||
182 | * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and | |||
183 | * let rhat = (u[j]*B + u[j+1]) mod v[1]. | |||
184 | * While rhat < B and v[2]*qhat > rhat*B+u[j+2], | |||
185 | * decrement qhat and increase rhat correspondingly. | |||
186 | * Note that if rhat >= B, v[2]*qhat < rhat*B. | |||
187 | */ | |||
188 | uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */ | |||
189 | uj1 = u[j + 1]; /* for D3 only */ | |||
190 | uj2 = u[j + 2]; /* for D3 only */ | |||
191 | if (uj0 == v1) { | |||
192 | qhat = B((int)1 << (sizeof(int) * 8 / 2)); | |||
193 | rhat = uj1; | |||
194 | goto qhat_too_big; | |||
195 | } else { | |||
196 | u_int nn = COMBINE(uj0, uj1)(((u_int)(uj0) << (sizeof(int) * 8 / 2)) | (uj1)); | |||
197 | qhat = nn / v1; | |||
198 | rhat = nn % v1; | |||
199 | } | |||
200 | while (v2 * qhat > COMBINE(rhat, uj2)(((u_int)(rhat) << (sizeof(int) * 8 / 2)) | (uj2))) { | |||
201 | qhat_too_big: | |||
202 | qhat--; | |||
203 | if ((rhat += v1) >= B((int)1 << (sizeof(int) * 8 / 2))) | |||
204 | break; | |||
205 | } | |||
206 | /* | |||
207 | * D4: Multiply and subtract. | |||
208 | * The variable `t' holds any borrows across the loop. | |||
209 | * We split this up so that we do not require v[0] = 0, | |||
210 | * and to eliminate a final special case. | |||
211 | */ | |||
212 | for (t = 0, i = n; i > 0; i--) { | |||
213 | t = u[i + j] - v[i] * qhat - t; | |||
214 | u[i + j] = (digit)LHALF(t)((u_int)(t) & (((int)1 << (sizeof(int) * 8 / 2)) - 1 )); | |||
215 | t = (B((int)1 << (sizeof(int) * 8 / 2)) - HHALF(t)((u_int)(t) >> (sizeof(int) * 8 / 2))) & (B((int)1 << (sizeof(int) * 8 / 2)) - 1); | |||
216 | } | |||
217 | t = u[j] - t; | |||
218 | u[j] = (digit)LHALF(t)((u_int)(t) & (((int)1 << (sizeof(int) * 8 / 2)) - 1 )); | |||
219 | /* | |||
220 | * D5: test remainder. | |||
221 | * There is a borrow if and only if HHALF(t) is nonzero; | |||
222 | * in that (rare) case, qhat was too large (by exactly 1). | |||
223 | * Fix it by adding v[1..n] to u[j..j+n]. | |||
224 | */ | |||
225 | if (HHALF(t)((u_int)(t) >> (sizeof(int) * 8 / 2))) { | |||
226 | qhat--; | |||
227 | for (t = 0, i = n; i > 0; i--) { /* D6: add back. */ | |||
228 | t += u[i + j] + v[i]; | |||
229 | u[i + j] = (digit)LHALF(t)((u_int)(t) & (((int)1 << (sizeof(int) * 8 / 2)) - 1 )); | |||
230 | t = HHALF(t)((u_int)(t) >> (sizeof(int) * 8 / 2)); | |||
231 | } | |||
232 | u[j] = (digit)LHALF(u[j] + t)((u_int)(u[j] + t) & (((int)1 << (sizeof(int) * 8 / 2)) - 1)); | |||
233 | } | |||
234 | q[j] = (digit)qhat; | |||
235 | } while (++j <= m); /* D7: loop on j. */ | |||
236 | ||||
237 | /* | |||
238 | * If caller wants the remainder, we have to calculate it as | |||
239 | * u[m..m+n] >> d (this is at most n digits and thus fits in | |||
240 | * u[m+1..m+n], but we may need more source digits). | |||
241 | */ | |||
242 | if (arq) { | |||
243 | if (d) { | |||
244 | for (i = m + n; i > m; --i) | |||
245 | u[i] = (digit)(((u_int)u[i] >> d) | | |||
246 | LHALF((u_int)u[i - 1] << (HALF_BITS - d))((u_int)((u_int)u[i - 1] << ((sizeof(int) * 8 / 2) - d) ) & (((int)1 << (sizeof(int) * 8 / 2)) - 1))); | |||
247 | u[i] = 0; | |||
248 | } | |||
249 | tmp.ul[H1] = COMBINE(uspace[1], uspace[2])(((u_int)(uspace[1]) << (sizeof(int) * 8 / 2)) | (uspace [2])); | |||
250 | tmp.ul[L0] = COMBINE(uspace[3], uspace[4])(((u_int)(uspace[3]) << (sizeof(int) * 8 / 2)) | (uspace [4])); | |||
251 | *arq = tmp.q; | |||
252 | } | |||
253 | ||||
254 | tmp.ul[H1] = COMBINE(qspace[1], qspace[2])(((u_int)(qspace[1]) << (sizeof(int) * 8 / 2)) | (qspace [2])); | |||
255 | tmp.ul[L0] = COMBINE(qspace[3], qspace[4])(((u_int)(qspace[3]) << (sizeof(int) * 8 / 2)) | (qspace [4])); | |||
256 | return (tmp.q); | |||
257 | } | |||
258 | ||||
259 | /* | |||
260 | * Shift p[0]..p[len] left `sh' bits, ignoring any bits that | |||
261 | * `fall out' the left (there never will be any such anyway). | |||
262 | * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS. | |||
263 | */ | |||
264 | static void | |||
265 | shl(digit *p, int len, int sh) | |||
266 | { | |||
267 | int i; | |||
268 | ||||
269 | for (i = 0; i < len; i++) | |||
270 | p[i] = (digit)(LHALF((u_int)p[i] << sh)((u_int)((u_int)p[i] << sh) & (((int)1 << (sizeof (int) * 8 / 2)) - 1)) | | |||
271 | ((u_int)p[i + 1] >> (HALF_BITS(sizeof(int) * 8 / 2) - sh))); | |||
272 | p[i] = (digit)(LHALF((u_int)p[i] << sh)((u_int)((u_int)p[i] << sh) & (((int)1 << (sizeof (int) * 8 / 2)) - 1))); | |||
273 | } |