1 /*
2 * Copyright 1995-2023 The OpenSSL Project Authors. All Rights Reserved.
3 *
4 * Licensed under the Apache License 2.0 (the "License"). You may not use
5 * this file except in compliance with the License. You can obtain a copy
6 * in the file LICENSE in the source distribution or at
7 * https://www.openssl.org/source/license.html
8 */
9
10 #include "internal/cryptlib.h"
11 #include "bn_local.h"
12
BN_RECP_CTX_init(BN_RECP_CTX * recp)13 void BN_RECP_CTX_init(BN_RECP_CTX *recp)
14 {
15 memset(recp, 0, sizeof(*recp));
16 bn_init(&(recp->N));
17 bn_init(&(recp->Nr));
18 }
19
BN_RECP_CTX_new(void)20 BN_RECP_CTX *BN_RECP_CTX_new(void)
21 {
22 BN_RECP_CTX *ret;
23
24 if ((ret = OPENSSL_zalloc(sizeof(*ret))) == NULL)
25 return NULL;
26
27 bn_init(&(ret->N));
28 bn_init(&(ret->Nr));
29 ret->flags = BN_FLG_MALLOCED;
30 return ret;
31 }
32
BN_RECP_CTX_free(BN_RECP_CTX * recp)33 void BN_RECP_CTX_free(BN_RECP_CTX *recp)
34 {
35 if (recp == NULL)
36 return;
37 BN_free(&recp->N);
38 BN_free(&recp->Nr);
39 if (recp->flags & BN_FLG_MALLOCED)
40 OPENSSL_free(recp);
41 }
42
BN_RECP_CTX_set(BN_RECP_CTX * recp,const BIGNUM * d,BN_CTX * ctx)43 int BN_RECP_CTX_set(BN_RECP_CTX *recp, const BIGNUM *d, BN_CTX *ctx)
44 {
45 if (BN_is_zero(d) || !BN_copy(&(recp->N), d))
46 return 0;
47 BN_zero(&(recp->Nr));
48 recp->num_bits = BN_num_bits(d);
49 recp->shift = 0;
50 return 1;
51 }
52
BN_mod_mul_reciprocal(BIGNUM * r,const BIGNUM * x,const BIGNUM * y,BN_RECP_CTX * recp,BN_CTX * ctx)53 int BN_mod_mul_reciprocal(BIGNUM *r, const BIGNUM *x, const BIGNUM *y,
54 BN_RECP_CTX *recp, BN_CTX *ctx)
55 {
56 int ret = 0;
57 BIGNUM *a;
58 const BIGNUM *ca;
59
60 BN_CTX_start(ctx);
61 if ((a = BN_CTX_get(ctx)) == NULL)
62 goto err;
63 if (y != NULL) {
64 if (x == y) {
65 if (!BN_sqr(a, x, ctx))
66 goto err;
67 } else {
68 if (!BN_mul(a, x, y, ctx))
69 goto err;
70 }
71 ca = a;
72 } else
73 ca = x; /* Just do the mod */
74
75 ret = BN_div_recp(NULL, r, ca, recp, ctx);
76 err:
77 BN_CTX_end(ctx);
78 bn_check_top(r);
79 return ret;
80 }
81
BN_div_recp(BIGNUM * dv,BIGNUM * rem,const BIGNUM * m,BN_RECP_CTX * recp,BN_CTX * ctx)82 int BN_div_recp(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m,
83 BN_RECP_CTX *recp, BN_CTX *ctx)
84 {
85 int i, j, ret = 0;
86 BIGNUM *a, *b, *d, *r;
87
88 BN_CTX_start(ctx);
89 d = (dv != NULL) ? dv : BN_CTX_get(ctx);
90 r = (rem != NULL) ? rem : BN_CTX_get(ctx);
91 a = BN_CTX_get(ctx);
92 b = BN_CTX_get(ctx);
93 if (b == NULL)
94 goto err;
95
96 if (BN_ucmp(m, &(recp->N)) < 0) {
97 BN_zero(d);
98 if (!BN_copy(r, m)) {
99 BN_CTX_end(ctx);
100 return 0;
101 }
102 BN_CTX_end(ctx);
103 return 1;
104 }
105
106 /*
107 * We want the remainder Given input of ABCDEF / ab we need multiply
108 * ABCDEF by 3 digests of the reciprocal of ab
109 */
110
111 /* i := max(BN_num_bits(m), 2*BN_num_bits(N)) */
112 i = BN_num_bits(m);
113 j = recp->num_bits << 1;
114 if (j > i)
115 i = j;
116
117 /* Nr := round(2^i / N) */
118 if (i != recp->shift)
119 recp->shift = BN_reciprocal(&(recp->Nr), &(recp->N), i, ctx);
120 /* BN_reciprocal could have returned -1 for an error */
121 if (recp->shift == -1)
122 goto err;
123
124 /*-
125 * d := |round(round(m / 2^BN_num_bits(N)) * recp->Nr / 2^(i - BN_num_bits(N)))|
126 * = |round(round(m / 2^BN_num_bits(N)) * round(2^i / N) / 2^(i - BN_num_bits(N)))|
127 * <= |(m / 2^BN_num_bits(N)) * (2^i / N) * (2^BN_num_bits(N) / 2^i)|
128 * = |m/N|
129 */
130 if (!BN_rshift(a, m, recp->num_bits))
131 goto err;
132 if (!BN_mul(b, a, &(recp->Nr), ctx))
133 goto err;
134 if (!BN_rshift(d, b, i - recp->num_bits))
135 goto err;
136 d->neg = 0;
137
138 if (!BN_mul(b, &(recp->N), d, ctx))
139 goto err;
140 if (!BN_usub(r, m, b))
141 goto err;
142 r->neg = 0;
143
144 j = 0;
145 while (BN_ucmp(r, &(recp->N)) >= 0) {
146 if (j++ > 2) {
147 ERR_raise(ERR_LIB_BN, BN_R_BAD_RECIPROCAL);
148 goto err;
149 }
150 if (!BN_usub(r, r, &(recp->N)))
151 goto err;
152 if (!BN_add_word(d, 1))
153 goto err;
154 }
155
156 r->neg = BN_is_zero(r) ? 0 : m->neg;
157 d->neg = m->neg ^ recp->N.neg;
158 ret = 1;
159 err:
160 BN_CTX_end(ctx);
161 bn_check_top(dv);
162 bn_check_top(rem);
163 return ret;
164 }
165
166 /*
167 * len is the expected size of the result We actually calculate with an extra
168 * word of precision, so we can do faster division if the remainder is not
169 * required.
170 */
171 /* r := 2^len / m */
BN_reciprocal(BIGNUM * r,const BIGNUM * m,int len,BN_CTX * ctx)172 int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx)
173 {
174 int ret = -1;
175 BIGNUM *t;
176
177 BN_CTX_start(ctx);
178 if ((t = BN_CTX_get(ctx)) == NULL)
179 goto err;
180
181 if (!BN_set_bit(t, len))
182 goto err;
183
184 if (!BN_div(r, NULL, t, m, ctx))
185 goto err;
186
187 ret = len;
188 err:
189 bn_check_top(r);
190 BN_CTX_end(ctx);
191 return ret;
192 }
193
