1 | /* |
2 | * Copyright 1995-2018 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 | |
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 | |
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 | BNerr(BN_F_BN_RECP_CTX_NEW, ERR_R_MALLOC_FAILURE); |
26 | return NULL; |
27 | } |
28 | |
29 | bn_init(&(ret->N)); |
30 | bn_init(&(ret->Nr)); |
31 | ret->flags = BN_FLG_MALLOCED; |
32 | return ret; |
33 | } |
34 | |
35 | void BN_RECP_CTX_free(BN_RECP_CTX *recp) |
36 | { |
37 | if (recp == NULL) |
38 | return; |
39 | BN_free(&recp->N); |
40 | BN_free(&recp->Nr); |
41 | if (recp->flags & BN_FLG_MALLOCED) |
42 | OPENSSL_free(recp); |
43 | } |
44 | |
45 | int BN_RECP_CTX_set(BN_RECP_CTX *recp, const BIGNUM *d, BN_CTX *ctx) |
46 | { |
47 | if (!BN_copy(&(recp->N), d)) |
48 | return 0; |
49 | BN_zero(&(recp->Nr)); |
50 | recp->num_bits = BN_num_bits(d); |
51 | recp->shift = 0; |
52 | return 1; |
53 | } |
54 | |
55 | int BN_mod_mul_reciprocal(BIGNUM *r, const BIGNUM *x, const BIGNUM *y, |
56 | BN_RECP_CTX *recp, BN_CTX *ctx) |
57 | { |
58 | int ret = 0; |
59 | BIGNUM *a; |
60 | const BIGNUM *ca; |
61 | |
62 | BN_CTX_start(ctx); |
63 | if ((a = BN_CTX_get(ctx)) == NULL) |
64 | goto err; |
65 | if (y != NULL) { |
66 | if (x == y) { |
67 | if (!BN_sqr(a, x, ctx)) |
68 | goto err; |
69 | } else { |
70 | if (!BN_mul(a, x, y, ctx)) |
71 | goto err; |
72 | } |
73 | ca = a; |
74 | } else |
75 | ca = x; /* Just do the mod */ |
76 | |
77 | ret = BN_div_recp(NULL, r, ca, recp, ctx); |
78 | err: |
79 | BN_CTX_end(ctx); |
80 | bn_check_top(r); |
81 | return ret; |
82 | } |
83 | |
84 | int BN_div_recp(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m, |
85 | BN_RECP_CTX *recp, BN_CTX *ctx) |
86 | { |
87 | int i, j, ret = 0; |
88 | BIGNUM *a, *b, *d, *r; |
89 | |
90 | BN_CTX_start(ctx); |
91 | d = (dv != NULL) ? dv : BN_CTX_get(ctx); |
92 | r = (rem != NULL) ? rem : BN_CTX_get(ctx); |
93 | a = BN_CTX_get(ctx); |
94 | b = BN_CTX_get(ctx); |
95 | if (b == NULL) |
96 | goto err; |
97 | |
98 | if (BN_ucmp(m, &(recp->N)) < 0) { |
99 | BN_zero(d); |
100 | if (!BN_copy(r, m)) { |
101 | BN_CTX_end(ctx); |
102 | return 0; |
103 | } |
104 | BN_CTX_end(ctx); |
105 | return 1; |
106 | } |
107 | |
108 | /* |
109 | * We want the remainder Given input of ABCDEF / ab we need multiply |
110 | * ABCDEF by 3 digests of the reciprocal of ab |
111 | */ |
112 | |
113 | /* i := max(BN_num_bits(m), 2*BN_num_bits(N)) */ |
114 | i = BN_num_bits(m); |
115 | j = recp->num_bits << 1; |
116 | if (j > i) |
117 | i = j; |
118 | |
119 | /* Nr := round(2^i / N) */ |
120 | if (i != recp->shift) |
121 | recp->shift = BN_reciprocal(&(recp->Nr), &(recp->N), i, ctx); |
122 | /* BN_reciprocal could have returned -1 for an error */ |
123 | if (recp->shift == -1) |
124 | goto err; |
125 | |
126 | /*- |
127 | * d := |round(round(m / 2^BN_num_bits(N)) * recp->Nr / 2^(i - BN_num_bits(N)))| |
128 | * = |round(round(m / 2^BN_num_bits(N)) * round(2^i / N) / 2^(i - BN_num_bits(N)))| |
129 | * <= |(m / 2^BN_num_bits(N)) * (2^i / N) * (2^BN_num_bits(N) / 2^i)| |
130 | * = |m/N| |
131 | */ |
132 | if (!BN_rshift(a, m, recp->num_bits)) |
133 | goto err; |
134 | if (!BN_mul(b, a, &(recp->Nr), ctx)) |
135 | goto err; |
136 | if (!BN_rshift(d, b, i - recp->num_bits)) |
137 | goto err; |
138 | d->neg = 0; |
139 | |
140 | if (!BN_mul(b, &(recp->N), d, ctx)) |
141 | goto err; |
142 | if (!BN_usub(r, m, b)) |
143 | goto err; |
144 | r->neg = 0; |
145 | |
146 | j = 0; |
147 | while (BN_ucmp(r, &(recp->N)) >= 0) { |
148 | if (j++ > 2) { |
149 | BNerr(BN_F_BN_DIV_RECP, BN_R_BAD_RECIPROCAL); |
150 | goto err; |
151 | } |
152 | if (!BN_usub(r, r, &(recp->N))) |
153 | goto err; |
154 | if (!BN_add_word(d, 1)) |
155 | goto err; |
156 | } |
157 | |
158 | r->neg = BN_is_zero(r) ? 0 : m->neg; |
159 | d->neg = m->neg ^ recp->N.neg; |
160 | ret = 1; |
161 | err: |
162 | BN_CTX_end(ctx); |
163 | bn_check_top(dv); |
164 | bn_check_top(rem); |
165 | return ret; |
166 | } |
167 | |
168 | /* |
169 | * len is the expected size of the result We actually calculate with an extra |
170 | * word of precision, so we can do faster division if the remainder is not |
171 | * required. |
172 | */ |
173 | /* r := 2^len / m */ |
174 | int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx) |
175 | { |
176 | int ret = -1; |
177 | BIGNUM *t; |
178 | |
179 | BN_CTX_start(ctx); |
180 | if ((t = BN_CTX_get(ctx)) == NULL) |
181 | goto err; |
182 | |
183 | if (!BN_set_bit(t, len)) |
184 | goto err; |
185 | |
186 | if (!BN_div(r, NULL, t, m, ctx)) |
187 | goto err; |
188 | |
189 | ret = len; |
190 | err: |
191 | bn_check_top(r); |
192 | BN_CTX_end(ctx); |
193 | return ret; |
194 | } |
195 | |