/* * Copyright 1995-2018 The OpenSSL Project Authors. All Rights Reserved. * * Licensed under the OpenSSL license (the "License"). You may not use * this file except in compliance with the License. You can obtain a copy * in the file LICENSE in the source distribution or at * https://www.openssl.org/source/license.html */ #include "internal/cryptlib.h" #include "bn_local.h" void BN_RECP_CTX_init(BN_RECP_CTX *recp) { memset(recp, 0, sizeof(*recp)); bn_init(&(recp->N)); bn_init(&(recp->Nr)); } BN_RECP_CTX *BN_RECP_CTX_new(void) { BN_RECP_CTX *ret; if ((ret = OPENSSL_zalloc(sizeof(*ret))) == NULL) { BNerr(BN_F_BN_RECP_CTX_NEW, ERR_R_MALLOC_FAILURE); return NULL; } bn_init(&(ret->N)); bn_init(&(ret->Nr)); ret->flags = BN_FLG_MALLOCED; return ret; } void BN_RECP_CTX_free(BN_RECP_CTX *recp) { if (recp == NULL) return; BN_free(&recp->N); BN_free(&recp->Nr); if (recp->flags & BN_FLG_MALLOCED) OPENSSL_free(recp); } int BN_RECP_CTX_set(BN_RECP_CTX *recp, const BIGNUM *d, BN_CTX *ctx) { if (!BN_copy(&(recp->N), d)) return 0; BN_zero(&(recp->Nr)); recp->num_bits = BN_num_bits(d); recp->shift = 0; return 1; } int BN_mod_mul_reciprocal(BIGNUM *r, const BIGNUM *x, const BIGNUM *y, BN_RECP_CTX *recp, BN_CTX *ctx) { int ret = 0; BIGNUM *a; const BIGNUM *ca; BN_CTX_start(ctx); if ((a = BN_CTX_get(ctx)) == NULL) goto err; if (y != NULL) { if (x == y) { if (!BN_sqr(a, x, ctx)) goto err; } else { if (!BN_mul(a, x, y, ctx)) goto err; } ca = a; } else ca = x; /* Just do the mod */ ret = BN_div_recp(NULL, r, ca, recp, ctx); err: BN_CTX_end(ctx); bn_check_top(r); return ret; } int BN_div_recp(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m, BN_RECP_CTX *recp, BN_CTX *ctx) { int i, j, ret = 0; BIGNUM *a, *b, *d, *r; BN_CTX_start(ctx); d = (dv != NULL) ? dv : BN_CTX_get(ctx); r = (rem != NULL) ? rem : BN_CTX_get(ctx); a = BN_CTX_get(ctx); b = BN_CTX_get(ctx); if (b == NULL) goto err; if (BN_ucmp(m, &(recp->N)) < 0) { BN_zero(d); if (!BN_copy(r, m)) { BN_CTX_end(ctx); return 0; } BN_CTX_end(ctx); return 1; } /* * We want the remainder Given input of ABCDEF / ab we need multiply * ABCDEF by 3 digests of the reciprocal of ab */ /* i := max(BN_num_bits(m), 2*BN_num_bits(N)) */ i = BN_num_bits(m); j = recp->num_bits << 1; if (j > i) i = j; /* Nr := round(2^i / N) */ if (i != recp->shift) recp->shift = BN_reciprocal(&(recp->Nr), &(recp->N), i, ctx); /* BN_reciprocal could have returned -1 for an error */ if (recp->shift == -1) goto err; /*- * d := |round(round(m / 2^BN_num_bits(N)) * recp->Nr / 2^(i - BN_num_bits(N)))| * = |round(round(m / 2^BN_num_bits(N)) * round(2^i / N) / 2^(i - BN_num_bits(N)))| * <= |(m / 2^BN_num_bits(N)) * (2^i / N) * (2^BN_num_bits(N) / 2^i)| * = |m/N| */ if (!BN_rshift(a, m, recp->num_bits)) goto err; if (!BN_mul(b, a, &(recp->Nr), ctx)) goto err; if (!BN_rshift(d, b, i - recp->num_bits)) goto err; d->neg = 0; if (!BN_mul(b, &(recp->N), d, ctx)) goto err; if (!BN_usub(r, m, b)) goto err; r->neg = 0; j = 0; while (BN_ucmp(r, &(recp->N)) >= 0) { if (j++ > 2) { BNerr(BN_F_BN_DIV_RECP, BN_R_BAD_RECIPROCAL); goto err; } if (!BN_usub(r, r, &(recp->N))) goto err; if (!BN_add_word(d, 1)) goto err; } r->neg = BN_is_zero(r) ? 0 : m->neg; d->neg = m->neg ^ recp->N.neg; ret = 1; err: BN_CTX_end(ctx); bn_check_top(dv); bn_check_top(rem); return ret; } /* * len is the expected size of the result We actually calculate with an extra * word of precision, so we can do faster division if the remainder is not * required. */ /* r := 2^len / m */ int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx) { int ret = -1; BIGNUM *t; BN_CTX_start(ctx); if ((t = BN_CTX_get(ctx)) == NULL) goto err; if (!BN_set_bit(t, len)) goto err; if (!BN_div(r, NULL, t, m, ctx)) goto err; ret = len; err: bn_check_top(r); BN_CTX_end(ctx); return ret; }