Bug Summary

File:src/sys/arch/amd64/stand/efiboot/bootia32/../../../../../lib/libkern/qdivrem.c
Warning:line 80, column 29
Division by zero

Annotated Source Code

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clang -cc1 -cc1 -triple i386-unknown-openbsd7.0 -analyze -disable-free -disable-llvm-verifier -discard-value-names -main-file-name qdivrem.c -analyzer-store=region -analyzer-opt-analyze-nested-blocks -analyzer-checker=core -analyzer-checker=apiModeling -analyzer-checker=unix -analyzer-checker=deadcode -analyzer-checker=security.insecureAPI.UncheckedReturn -analyzer-checker=security.insecureAPI.getpw -analyzer-checker=security.insecureAPI.gets -analyzer-checker=security.insecureAPI.mktemp -analyzer-checker=security.insecureAPI.mkstemp -analyzer-checker=security.insecureAPI.vfork -analyzer-checker=nullability.NullPassedToNonnull -analyzer-checker=nullability.NullReturnedFromNonnull -analyzer-output plist -w -setup-static-analyzer -mrelocation-model pic -pic-level 2 -fhalf-no-semantic-interposition -mframe-pointer=all -relaxed-aliasing -fno-rounding-math -mconstructor-aliases -ffreestanding -target-cpu i586 -disable-red-zone -tune-cpu generic -debugger-tuning=gdb -fcoverage-compilation-dir=/usr/src/sys/arch/amd64/stand/efiboot/bootia32/obj -nostdsysteminc -nobuiltininc -resource-dir /usr/local/lib/clang/13.0.0 -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../../../../stand/efi/include/i386 -D EFIBOOT -D FWRANDOM -D NEEDS_HEAP_H -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/.. -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../../../../stand/efi/include -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../../../../stand/boot -D SOFTRAID -D _STANDALONE -D MDRANDOM -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../../../../stand/efi/include/i386 -D EFIBOOT -D FWRANDOM -D NEEDS_HEAP_H -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/.. -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../../../../stand/efi/include -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../../../../stand/boot -D SOFTRAID -D _STANDALONE -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../../../.. -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../libsa -I . -I /usr/src/sys/arch/amd64/stand/efiboot/bootia32 -D SMALL -D SLOW -D NOBYFOUR -D __INTERNAL_LIBSA_CREAD -D HEAP_LIMIT=0xc00000 -D HIBERNATE -Oz -Wno-pointer-sign -std=gnu99 -fdebug-compilation-dir=/usr/src/sys/arch/amd64/stand/efiboot/bootia32/obj -ferror-limit 19 -fwrapv -fno-builtin -fwchar-type=short -fno-signed-wchar -fgnuc-version=4.2.1 -vectorize-slp -fno-builtin-malloc -fno-builtin-calloc -fno-builtin-realloc -fno-builtin-valloc -fno-builtin-free -fno-builtin-strdup -fno-builtin-strndup -analyzer-output=html -faddrsig -o /home/ben/Projects/vmm/scan-build/2022-01-12-194120-40624-1 -x c /usr/src/sys/arch/amd64/stand/efiboot/bootia32/../../../../../lib/libkern/qdivrem.c
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
48typedef unsigned short digit;
49#else
50typedef u_int digit;
51#endif
52
53static 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 */
63u_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) {
1
Assuming 'vq' is equal to 0
→
2
←
Taking true branch
→
77 /* divide by zero. */
78 static volatile const unsigned int zero = 0;
3
←
'zero' initialized to 0
→
79
80 tmp.ul[H1] = tmp.ul[L0] = 1 / zero;
4
←
Division by 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 */
264static void
265shl(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}