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/*
* Copyright 1995-2016 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_lcl.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)
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);
a = BN_CTX_get(ctx);
b = BN_CTX_get(ctx);
if (dv != NULL)
d = dv;
else
d = BN_CTX_get(ctx);
if (rem != NULL)
r = rem;
else
r = BN_CTX_get(ctx);
if (a == NULL || b == NULL || d == NULL || r == 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);
}
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