openssl/crypto/bn/bn_recp.c
Matt Caswell 320a81277e Remove some code for a contributor that we cannot find
This removes some code because we cannot trace the original contributor
to get their agreement for the licence change (original commit e03ddfae).

After this change there will be numerous failures in the test cases until
someone rewrites the missing code.

All *_free functions should accept a NULL parameter. After this change
the following *_free functions will fail if a NULL parameter is passed:

BIO_ACCEPT_free()
BIO_CONNECT_free()
BN_BLINDING_free()
BN_CTX_free()
BN_MONT_CTX_free()
BN_RECP_CTX_free()
BUF_MEM_free()
COMP_CTX_free()
ERR_STATE_free()
TXT_DB_free()
X509_STORE_free()
ssl3_free()
ssl_cert_free()
SSL_SESSION_free()
SSL_free()

[skip ci]

Reviewed-by: Bernd Edlinger <bernd.edlinger@hotmail.de>
(Merged from https://github.com/openssl/openssl/pull/5757)
2018-03-27 17:15:24 +01:00

190 lines
4.4 KiB
C

/*
* Copyright 1995-2017 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)
{
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;
}