edea42c602
To make it consistent in the code base Reviewed-by: Matt Caswell <matt@openssl.org> Reviewed-by: Bernd Edlinger <bernd.edlinger@hotmail.de> (Merged from https://github.com/openssl/openssl/pull/3749)
193 lines
4.5 KiB
C
193 lines
4.5 KiB
C
/*
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* Copyright 1995-2017 The OpenSSL Project Authors. All Rights Reserved.
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*
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* Licensed under the OpenSSL license (the "License"). You may not use
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* this file except in compliance with the License. You can obtain a copy
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* in the file LICENSE in the source distribution or at
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* https://www.openssl.org/source/license.html
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*/
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#include "internal/cryptlib.h"
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#include "bn_lcl.h"
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void BN_RECP_CTX_init(BN_RECP_CTX *recp)
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{
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memset(recp, 0, sizeof(*recp));
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bn_init(&(recp->N));
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bn_init(&(recp->Nr));
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}
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BN_RECP_CTX *BN_RECP_CTX_new(void)
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{
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BN_RECP_CTX *ret;
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if ((ret = OPENSSL_zalloc(sizeof(*ret))) == NULL)
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return (NULL);
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bn_init(&(ret->N));
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bn_init(&(ret->Nr));
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ret->flags = BN_FLG_MALLOCED;
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return (ret);
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}
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void BN_RECP_CTX_free(BN_RECP_CTX *recp)
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{
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if (recp == NULL)
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return;
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BN_free(&(recp->N));
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BN_free(&(recp->Nr));
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if (recp->flags & BN_FLG_MALLOCED)
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OPENSSL_free(recp);
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}
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int BN_RECP_CTX_set(BN_RECP_CTX *recp, const BIGNUM *d, BN_CTX *ctx)
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{
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if (!BN_copy(&(recp->N), d))
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return 0;
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BN_zero(&(recp->Nr));
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recp->num_bits = BN_num_bits(d);
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recp->shift = 0;
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return (1);
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}
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int BN_mod_mul_reciprocal(BIGNUM *r, const BIGNUM *x, const BIGNUM *y,
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BN_RECP_CTX *recp, BN_CTX *ctx)
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{
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int ret = 0;
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BIGNUM *a;
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const BIGNUM *ca;
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BN_CTX_start(ctx);
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if ((a = BN_CTX_get(ctx)) == NULL)
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goto err;
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if (y != NULL) {
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if (x == y) {
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if (!BN_sqr(a, x, ctx))
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goto err;
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} else {
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if (!BN_mul(a, x, y, ctx))
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goto err;
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}
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ca = a;
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} else
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ca = x; /* Just do the mod */
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ret = BN_div_recp(NULL, r, ca, recp, ctx);
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err:
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BN_CTX_end(ctx);
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bn_check_top(r);
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return (ret);
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}
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int BN_div_recp(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m,
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BN_RECP_CTX *recp, BN_CTX *ctx)
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{
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int i, j, ret = 0;
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BIGNUM *a, *b, *d, *r;
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BN_CTX_start(ctx);
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d = (dv != NULL) ? dv : BN_CTX_get(ctx);
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r = (rem != NULL) ? rem : BN_CTX_get(ctx);
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a = BN_CTX_get(ctx);
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b = BN_CTX_get(ctx);
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if (b == NULL)
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goto err;
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if (BN_ucmp(m, &(recp->N)) < 0) {
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BN_zero(d);
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if (!BN_copy(r, m)) {
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BN_CTX_end(ctx);
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return 0;
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}
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BN_CTX_end(ctx);
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return (1);
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}
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/*
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* We want the remainder Given input of ABCDEF / ab we need multiply
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* ABCDEF by 3 digests of the reciprocal of ab
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*/
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/* i := max(BN_num_bits(m), 2*BN_num_bits(N)) */
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i = BN_num_bits(m);
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j = recp->num_bits << 1;
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if (j > i)
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i = j;
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/* Nr := round(2^i / N) */
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if (i != recp->shift)
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recp->shift = BN_reciprocal(&(recp->Nr), &(recp->N), i, ctx);
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/* BN_reciprocal could have returned -1 for an error */
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if (recp->shift == -1)
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goto err;
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/*-
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* d := |round(round(m / 2^BN_num_bits(N)) * recp->Nr / 2^(i - BN_num_bits(N)))|
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* = |round(round(m / 2^BN_num_bits(N)) * round(2^i / N) / 2^(i - BN_num_bits(N)))|
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* <= |(m / 2^BN_num_bits(N)) * (2^i / N) * (2^BN_num_bits(N) / 2^i)|
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* = |m/N|
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*/
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if (!BN_rshift(a, m, recp->num_bits))
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goto err;
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if (!BN_mul(b, a, &(recp->Nr), ctx))
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goto err;
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if (!BN_rshift(d, b, i - recp->num_bits))
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goto err;
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d->neg = 0;
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if (!BN_mul(b, &(recp->N), d, ctx))
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goto err;
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if (!BN_usub(r, m, b))
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goto err;
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r->neg = 0;
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j = 0;
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while (BN_ucmp(r, &(recp->N)) >= 0) {
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if (j++ > 2) {
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BNerr(BN_F_BN_DIV_RECP, BN_R_BAD_RECIPROCAL);
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goto err;
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}
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if (!BN_usub(r, r, &(recp->N)))
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goto err;
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if (!BN_add_word(d, 1))
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goto err;
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}
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r->neg = BN_is_zero(r) ? 0 : m->neg;
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d->neg = m->neg ^ recp->N.neg;
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ret = 1;
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err:
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BN_CTX_end(ctx);
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bn_check_top(dv);
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bn_check_top(rem);
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return (ret);
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}
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/*
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* len is the expected size of the result We actually calculate with an extra
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* word of precision, so we can do faster division if the remainder is not
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* required.
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*/
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/* r := 2^len / m */
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int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx)
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{
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int ret = -1;
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BIGNUM *t;
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BN_CTX_start(ctx);
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if ((t = BN_CTX_get(ctx)) == NULL)
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goto err;
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if (!BN_set_bit(t, len))
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goto err;
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if (!BN_div(r, NULL, t, m, ctx))
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goto err;
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ret = len;
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err:
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bn_check_top(r);
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BN_CTX_end(ctx);
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return (ret);
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}
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