1/* mpz_and -- Logical and. 2 3Copyright 1991, 1993, 1994, 1996, 1997, 2000, 2001, 2003, 2005, 2012, 42015-2018 Free Software Foundation, Inc. 5 6This file is part of the GNU MP Library. 7 8The GNU MP Library is free software; you can redistribute it and/or modify 9it 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 15or 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 21or both in parallel, as here. 22 23The GNU MP Library is distributed in the hope that it will be useful, but 24WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY 25or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License 26for more details. 27 28You should have received copies of the GNU General Public License and the 29GNU Lesser General Public License along with the GNU MP Library. If not, 30see https://www.gnu.org/licenses/. */ 31 32#include "gmp-impl.h" 33 34void 35mpz_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