[331] | 1 | /*
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| 2 | * Copyright 1995-2016 The OpenSSL Project Authors. All Rights Reserved.
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| 3 | *
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| 4 | * Licensed under the OpenSSL license (the "License"). You may not use
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| 5 | * this file except in compliance with the License. You can obtain a copy
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| 6 | * in the file LICENSE in the source distribution or at
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| 7 | * https://www.openssl.org/source/license.html
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| 8 | */
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| 9 |
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| 10 | #include "internal/cryptlib.h"
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| 11 | #include "bn_lcl.h"
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| 12 |
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| 13 | void BN_RECP_CTX_init(BN_RECP_CTX *recp)
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| 14 | {
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| 15 | memset(recp, 0, sizeof(*recp));
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| 16 | bn_init(&(recp->N));
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| 17 | bn_init(&(recp->Nr));
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| 18 | }
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| 19 |
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| 20 | BN_RECP_CTX *BN_RECP_CTX_new(void)
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| 21 | {
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| 22 | BN_RECP_CTX *ret;
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| 23 |
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| 24 | if ((ret = OPENSSL_zalloc(sizeof(*ret))) == NULL)
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| 25 | return (NULL);
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| 26 |
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| 27 | bn_init(&(ret->N));
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| 28 | bn_init(&(ret->Nr));
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| 29 | ret->flags = BN_FLG_MALLOCED;
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| 30 | return (ret);
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| 31 | }
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| 32 |
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| 33 | void BN_RECP_CTX_free(BN_RECP_CTX *recp)
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| 34 | {
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| 35 | if (recp == NULL)
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| 36 | return;
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| 37 |
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| 38 | BN_free(&(recp->N));
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| 39 | BN_free(&(recp->Nr));
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| 40 | if (recp->flags & BN_FLG_MALLOCED)
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| 41 | OPENSSL_free(recp);
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| 42 | }
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| 43 |
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| 44 | int BN_RECP_CTX_set(BN_RECP_CTX *recp, const BIGNUM *d, BN_CTX *ctx)
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| 45 | {
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| 46 | if (!BN_copy(&(recp->N), d))
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| 47 | return 0;
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| 48 | BN_zero(&(recp->Nr));
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| 49 | recp->num_bits = BN_num_bits(d);
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| 50 | recp->shift = 0;
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| 51 | return (1);
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| 52 | }
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| 53 |
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| 54 | int BN_mod_mul_reciprocal(BIGNUM *r, const BIGNUM *x, const BIGNUM *y,
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| 55 | BN_RECP_CTX *recp, BN_CTX *ctx)
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| 56 | {
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| 57 | int ret = 0;
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| 58 | BIGNUM *a;
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| 59 | const BIGNUM *ca;
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| 60 |
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| 61 | BN_CTX_start(ctx);
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| 62 | if ((a = BN_CTX_get(ctx)) == NULL)
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| 63 | goto err;
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| 64 | if (y != NULL) {
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| 65 | if (x == y) {
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| 66 | if (!BN_sqr(a, x, ctx))
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| 67 | goto err;
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| 68 | } else {
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| 69 | if (!BN_mul(a, x, y, ctx))
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| 70 | goto err;
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| 71 | }
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| 72 | ca = a;
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| 73 | } else
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| 74 | ca = x; /* Just do the mod */
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| 75 |
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| 76 | ret = BN_div_recp(NULL, r, ca, recp, ctx);
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| 77 | err:
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| 78 | BN_CTX_end(ctx);
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| 79 | bn_check_top(r);
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| 80 | return (ret);
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| 81 | }
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| 82 |
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| 83 | int BN_div_recp(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m,
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| 84 | BN_RECP_CTX *recp, BN_CTX *ctx)
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| 85 | {
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| 86 | int i, j, ret = 0;
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| 87 | BIGNUM *a, *b, *d, *r;
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| 88 |
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| 89 | BN_CTX_start(ctx);
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| 90 | a = BN_CTX_get(ctx);
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| 91 | b = BN_CTX_get(ctx);
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| 92 | if (dv != NULL)
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| 93 | d = dv;
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| 94 | else
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| 95 | d = BN_CTX_get(ctx);
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| 96 | if (rem != NULL)
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| 97 | r = rem;
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| 98 | else
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| 99 | r = BN_CTX_get(ctx);
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| 100 | if (a == NULL || b == NULL || d == NULL || r == NULL)
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| 101 | goto err;
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| 102 |
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| 103 | if (BN_ucmp(m, &(recp->N)) < 0) {
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| 104 | BN_zero(d);
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| 105 | if (!BN_copy(r, m)) {
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| 106 | BN_CTX_end(ctx);
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| 107 | return 0;
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| 108 | }
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| 109 | BN_CTX_end(ctx);
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| 110 | return (1);
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| 111 | }
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| 112 |
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| 113 | /*
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| 114 | * We want the remainder Given input of ABCDEF / ab we need multiply
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| 115 | * ABCDEF by 3 digests of the reciprocal of ab
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| 116 | */
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| 117 |
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| 118 | /* i := max(BN_num_bits(m), 2*BN_num_bits(N)) */
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| 119 | i = BN_num_bits(m);
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| 120 | j = recp->num_bits << 1;
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| 121 | if (j > i)
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| 122 | i = j;
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| 123 |
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| 124 | /* Nr := round(2^i / N) */
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| 125 | if (i != recp->shift)
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| 126 | recp->shift = BN_reciprocal(&(recp->Nr), &(recp->N), i, ctx);
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| 127 | /* BN_reciprocal could have returned -1 for an error */
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| 128 | if (recp->shift == -1)
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| 129 | goto err;
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| 130 |
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| 131 | /*-
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| 132 | * d := |round(round(m / 2^BN_num_bits(N)) * recp->Nr / 2^(i - BN_num_bits(N)))|
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| 133 | * = |round(round(m / 2^BN_num_bits(N)) * round(2^i / N) / 2^(i - BN_num_bits(N)))|
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| 134 | * <= |(m / 2^BN_num_bits(N)) * (2^i / N) * (2^BN_num_bits(N) / 2^i)|
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| 135 | * = |m/N|
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| 136 | */
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| 137 | if (!BN_rshift(a, m, recp->num_bits))
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| 138 | goto err;
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| 139 | if (!BN_mul(b, a, &(recp->Nr), ctx))
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| 140 | goto err;
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| 141 | if (!BN_rshift(d, b, i - recp->num_bits))
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| 142 | goto err;
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| 143 | d->neg = 0;
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| 144 |
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| 145 | if (!BN_mul(b, &(recp->N), d, ctx))
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| 146 | goto err;
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| 147 | if (!BN_usub(r, m, b))
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| 148 | goto err;
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| 149 | r->neg = 0;
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| 150 |
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| 151 | j = 0;
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| 152 | while (BN_ucmp(r, &(recp->N)) >= 0) {
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| 153 | if (j++ > 2) {
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| 154 | BNerr(BN_F_BN_DIV_RECP, BN_R_BAD_RECIPROCAL);
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| 155 | goto err;
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| 156 | }
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| 157 | if (!BN_usub(r, r, &(recp->N)))
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| 158 | goto err;
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| 159 | if (!BN_add_word(d, 1))
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| 160 | goto err;
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| 161 | }
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| 162 |
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| 163 | r->neg = BN_is_zero(r) ? 0 : m->neg;
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| 164 | d->neg = m->neg ^ recp->N.neg;
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| 165 | ret = 1;
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| 166 | err:
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| 167 | BN_CTX_end(ctx);
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| 168 | bn_check_top(dv);
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| 169 | bn_check_top(rem);
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| 170 | return (ret);
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| 171 | }
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| 172 |
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| 173 | /*
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| 174 | * len is the expected size of the result We actually calculate with an extra
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| 175 | * word of precision, so we can do faster division if the remainder is not
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| 176 | * required.
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| 177 | */
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| 178 | /* r := 2^len / m */
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| 179 | int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx)
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| 180 | {
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| 181 | int ret = -1;
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| 182 | BIGNUM *t;
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| 183 |
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| 184 | BN_CTX_start(ctx);
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| 185 | if ((t = BN_CTX_get(ctx)) == NULL)
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| 186 | goto err;
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| 187 |
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| 188 | if (!BN_set_bit(t, len))
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| 189 | goto err;
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| 190 |
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| 191 | if (!BN_div(r, NULL, t, m, ctx))
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| 192 | goto err;
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| 193 |
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| 194 | ret = len;
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| 195 | err:
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| 196 | bn_check_top(r);
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| 197 | BN_CTX_end(ctx);
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| 198 | return (ret);
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| 199 | }
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