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