1 /* mpz_and -- Logical and.
2
3 Copyright 1991, 1993, 1994, 1996, 1997, 2000, 2001, 2003, 2005, 2012,
4 2015-2018 Free Software Foundation, Inc.
5
6 This file is part of the GNU MP Library.
7
8 The GNU MP Library is free software; you can redistribute it and/or modify
9 it under the terms of either:
10
11 * the GNU Lesser General Public License as published by the Free
12 Software Foundation; either version 3 of the License, or (at your
13 option) any later version.
14
15 or
16
17 * the GNU General Public License as published by the Free Software
18 Foundation; either version 2 of the License, or (at your option) any
19 later version.
20
21 or both in parallel, as here.
22
23 The GNU MP Library is distributed in the hope that it will be useful, but
24 WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
25 or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
26 for more details.
27
28 You should have received copies of the GNU General Public License and the
29 GNU Lesser General Public License along with the GNU MP Library. If not,
30 see https://www.gnu.org/licenses/. */
31
32 #include "gmp-impl.h"
33
34 void
mpz_and(mpz_ptr res,mpz_srcptr op1,mpz_srcptr op2)35 mpz_and (mpz_ptr res, mpz_srcptr op1, mpz_srcptr op2)
36 {
37 mp_srcptr op1_ptr, op2_ptr;
38 mp_size_t op1_size, op2_size;
39 mp_ptr res_ptr;
40 mp_size_t res_size;
41 mp_size_t i;
42
43 op1_size = SIZ(op1);
44 op2_size = SIZ(op2);
45
46 if (op1_size < op2_size)
47 {
48 MPZ_SRCPTR_SWAP (op1, op2);
49 MP_SIZE_T_SWAP (op1_size, op2_size);
50 }
51
52 op1_ptr = PTR(op1);
53 op2_ptr = PTR(op2);
54
55 if (op2_size >= 0)
56 {
57 /* First loop finds the size of the result. */
58 for (i = op2_size; --i >= 0;)
59 if ((op1_ptr[i] & op2_ptr[i]) != 0)
60 {
61 res_size = i + 1;
62 /* Handle allocation, now then we know exactly how much space is
63 needed for the result. */
64 /* Don't re-read op1_ptr and op2_ptr. Since res_size <=
65 MIN(op1_size, op2_size), res is not changed when op1
66 is identical to res or op2 is identical to res. */
67 SIZ (res) = res_size;
68 mpn_and_n (MPZ_NEWALLOC (res, res_size), op1_ptr, op2_ptr, res_size);
69 return;
70 }
71
72 SIZ (res) = 0;
73 }
74 else
75 {
76 TMP_DECL;
77
78 op2_size = -op2_size;
79 TMP_MARK;
80 if (op1_size < 0)
81 {
82 mp_ptr opx, opy;
83
84 /* Both operands are negative, so will be the result.
85 -((-OP1) & (-OP2)) = -(~(OP1 - 1) & ~(OP2 - 1)) =
86 = ~(~(OP1 - 1) & ~(OP2 - 1)) + 1 =
87 = ((OP1 - 1) | (OP2 - 1)) + 1 */
88
89 /* It might seem as we could end up with an (invalid) result with
90 a leading zero-limb here when one of the operands is of the
91 type 1,,0,,..,,.0. But some analysis shows that we surely
92 would get carry into the zero-limb in this situation... */
93
94 op1_size = -op1_size;
95
96 TMP_ALLOC_LIMBS_2 (opx, op1_size, opy, op2_size);
97 mpn_sub_1 (opx, op1_ptr, op1_size, (mp_limb_t) 1);
98 op1_ptr = opx;
99
100 mpn_sub_1 (opy, op2_ptr, op2_size, (mp_limb_t) 1);
101 op2_ptr = opy;
102
103 res_ptr = MPZ_NEWALLOC (res, 1 + op2_size);
104 /* Don't re-read OP1_PTR and OP2_PTR. They point to temporary
105 space--never to the space PTR(res) used to point to before
106 reallocation. */
107
108 MPN_COPY (res_ptr + op1_size, op2_ptr + op1_size,
109 op2_size - op1_size);
110 mpn_ior_n (res_ptr, op1_ptr, op2_ptr, op1_size);
111 TMP_FREE;
112 res_size = op2_size;
113
114 res_ptr[res_size] = 0;
115 MPN_INCR_U (res_ptr, res_size + 1, (mp_limb_t) 1);
116 res_size += res_ptr[res_size];
117
118 SIZ(res) = -res_size;
119 }
120 else
121 {
122 #if ANDNEW
123 mp_size_t op2_lim;
124 mp_size_t count;
125
126 /* OP2 must be negated as with infinite precision.
127
128 Scan from the low end for a non-zero limb. The first non-zero
129 limb is simply negated (two's complement). Any subsequent
130 limbs are one's complemented. Of course, we don't need to
131 handle more limbs than there are limbs in the other, positive
132 operand as the result for those limbs is going to become zero
133 anyway. */
134
135 /* Scan for the least significant non-zero OP2 limb, and zero the
136 result meanwhile for those limb positions. (We will surely
137 find a non-zero limb, so we can write the loop with one
138 termination condition only.) */
139 for (i = 0; op2_ptr[i] == 0; i++)
140 res_ptr[i] = 0;
141 op2_lim = i;
142
143 if (op1_size <= op2_size)
144 {
145 /* The ones-extended OP2 is >= than the zero-extended OP1.
146 RES_SIZE <= OP1_SIZE. Find the exact size. */
147 for (i = op1_size - 1; i > op2_lim; i--)
148 if ((op1_ptr[i] & ~op2_ptr[i]) != 0)
149 break;
150 res_size = i + 1;
151 for (i = res_size - 1; i > op2_lim; i--)
152 res_ptr[i] = op1_ptr[i] & ~op2_ptr[i];
153 res_ptr[op2_lim] = op1_ptr[op2_lim] & -op2_ptr[op2_lim];
154 /* Yes, this *can* happen! */
155 MPN_NORMALIZE (res_ptr, res_size);
156 }
157 else
158 {
159 /* The ones-extended OP2 is < than the zero-extended OP1.
160 RES_SIZE == OP1_SIZE, since OP1 is normalized. */
161 res_size = op1_size;
162 MPN_COPY (res_ptr + op2_size, op1_ptr + op2_size, op1_size - op2_size);
163 for (i = op2_size - 1; i > op2_lim; i--)
164 res_ptr[i] = op1_ptr[i] & ~op2_ptr[i];
165 res_ptr[op2_lim] = op1_ptr[op2_lim] & -op2_ptr[op2_lim];
166 }
167 #else
168
169 /* OP1 is positive and zero-extended,
170 OP2 is negative and ones-extended.
171 The result will be positive.
172 OP1 & -OP2 = OP1 & ~(OP2 - 1). */
173
174 mp_ptr opx;
175
176 opx = TMP_ALLOC_LIMBS (op2_size);
177 mpn_sub_1 (opx, op2_ptr, op2_size, (mp_limb_t) 1);
178 op2_ptr = opx;
179
180 if (op1_size > op2_size)
181 {
182 /* The result has the same size as OP1, since OP1 is normalized
183 and longer than the ones-extended OP2. */
184 res_size = op1_size;
185
186 /* Handle allocation, now then we know exactly how much space is
187 needed for the result. */
188 res_ptr = MPZ_NEWALLOC (res, res_size);
189 /* Don't re-read OP1_PTR or OP2_PTR. Since res_size = op1_size,
190 op1 is not changed if it is identical to res.
191 OP2_PTR points to temporary space. */
192
193 mpn_andn_n (res_ptr, op1_ptr, op2_ptr, op2_size);
194 MPN_COPY (res_ptr + op2_size, op1_ptr + op2_size, res_size - op2_size);
195 }
196 else
197 {
198 /* Find out the exact result size. Ignore the high limbs of OP2,
199 OP1 is zero-extended and would make the result zero. */
200 res_size = 0;
201 for (i = op1_size; --i >= 0;)
202 if ((op1_ptr[i] & ~op2_ptr[i]) != 0)
203 {
204 res_size = i + 1;
205 /* Handle allocation, now then we know exactly how much
206 space is needed for the result. */
207 /* Don't re-read OP1_PTR. Since res_size <= op1_size,
208 op1 is not changed if it is identical to res. Don't
209 re-read OP2_PTR. It points to temporary space--never
210 to the space PTR(res) used to point to before
211 reallocation. */
212 mpn_andn_n (MPZ_NEWALLOC (res, res_size), op1_ptr, op2_ptr, res_size);
213
214 break;
215 }
216 }
217 #endif
218 SIZ(res) = res_size;
219 TMP_FREE;
220 }
221 }
222 }
223