-/* crypto/bn/bn_recp.c */
-/* Copyright (C) 1995-1998 Eric Young (eay@cryptsoft.com)
- * All rights reserved.
+/*
+ * Copyright 1995-2016 The OpenSSL Project Authors. All Rights Reserved.
*
- * This package is an SSL implementation written
- * by Eric Young (eay@cryptsoft.com).
- * The implementation was written so as to conform with Netscapes SSL.
- *
- * This library is free for commercial and non-commercial use as long as
- * the following conditions are aheared to. The following conditions
- * apply to all code found in this distribution, be it the RC4, RSA,
- * lhash, DES, etc., code; not just the SSL code. The SSL documentation
- * included with this distribution is covered by the same copyright terms
- * except that the holder is Tim Hudson (tjh@cryptsoft.com).
- *
- * Copyright remains Eric Young's, and as such any Copyright notices in
- * the code are not to be removed.
- * If this package is used in a product, Eric Young should be given attribution
- * as the author of the parts of the library used.
- * This can be in the form of a textual message at program startup or
- * in documentation (online or textual) provided with the package.
- *
- * Redistribution and use in source and binary forms, with or without
- * modification, are permitted provided that the following conditions
- * are met:
- * 1. Redistributions of source code must retain the copyright
- * notice, this list of conditions and the following disclaimer.
- * 2. Redistributions in binary form must reproduce the above copyright
- * notice, this list of conditions and the following disclaimer in the
- * documentation and/or other materials provided with the distribution.
- * 3. All advertising materials mentioning features or use of this software
- * must display the following acknowledgement:
- * "This product includes cryptographic software written by
- * Eric Young (eay@cryptsoft.com)"
- * The word 'cryptographic' can be left out if the rouines from the library
- * being used are not cryptographic related :-).
- * 4. If you include any Windows specific code (or a derivative thereof) from
- * the apps directory (application code) you must include an acknowledgement:
- * "This product includes software written by Tim Hudson (tjh@cryptsoft.com)"
- *
- * THIS SOFTWARE IS PROVIDED BY ERIC YOUNG ``AS IS'' AND
- * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
- * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
- * ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
- * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
- * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
- * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
- * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
- * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
- * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
- * SUCH DAMAGE.
- *
- * The licence and distribution terms for any publically available version or
- * derivative of this code cannot be changed. i.e. this code cannot simply be
- * copied and put under another distribution licence
- * [including the GNU Public Licence.]
+ * 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 <stdio.h>
-#include "cryptlib.h"
+#include "internal/cryptlib.h"
#include "bn_lcl.h"
void BN_RECP_CTX_init(BN_RECP_CTX *recp)
- {
- BN_init(&(recp->N));
- BN_init(&(recp->Nr));
- recp->num_bits=0;
- recp->flags=0;
- }
+{
+ 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;
+{
+ BN_RECP_CTX *ret;
- if ((ret=(BN_RECP_CTX *)Malloc(sizeof(BN_RECP_CTX))) == NULL)
- return(NULL);
+ if ((ret = OPENSSL_zalloc(sizeof(*ret))) == NULL)
+ return (NULL);
- BN_RECP_CTX_init(ret);
- ret->flags=BN_FLG_MALLOCED;
- return(ret);
- }
+ 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;
+{
+ if (recp == NULL)
+ return;
- BN_free(&(recp->N));
- BN_free(&(recp->Nr));
- if (recp->flags & BN_FLG_MALLOCED)
- Free(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)
- {
- BN_copy(&(recp->N),d);
- BN_zero(&(recp->Nr));
- recp->num_bits=BN_num_bits(d);
- recp->shift=0;
- return(1);
- }
-
-int BN_mod_mul_reciprocal(BIGNUM *r, BIGNUM *x, BIGNUM *y, BN_RECP_CTX *recp,
- BN_CTX *ctx)
- {
- int ret=0;
- BIGNUM *a;
-
- 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; }
- }
- else
- a=x; /* Just do the mod */
-
- BN_div_recp(NULL,r,a,recp,ctx);
- ret=1;
-err:
- BN_CTX_end(ctx);
- return(ret);
- }
-
-int BN_div_recp(BIGNUM *dv, BIGNUM *rem, BIGNUM *m, BN_RECP_CTX *recp,
- BN_CTX *ctx)
- {
- int i,j,ret=0,ex;
- 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);
- BN_copy(r,m);
- 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=BN_num_bits(m);
- if (i%2) i--;
-
- j=recp->num_bits*2;
- if (j > i)
- {
- i=j;
- ex=0;
- }
- else
- {
- ex=(i-j)/2;
- }
-
- j=i/2;
-
- if (i != recp->shift)
- recp->shift=BN_reciprocal(&(recp->Nr),&(recp->N),
- i,ctx);
-
- if (!BN_rshift(a,m,j-ex)) goto err;
- if (!BN_mul(b,a,&(recp->Nr),ctx)) goto err;
- if (!BN_rshift(d,b,j+ex)) 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;
-#if 1
- while (BN_ucmp(r,&(recp->N)) >= 0)
- {
- if (j++ > 2)
- {
- BNerr(BN_F_BN_MOD_MUL_RECIPROCAL,BN_R_BAD_RECIPROCAL);
- goto err;
- }
- if (!BN_usub(r,r,&(recp->N))) goto err;
- if (!BN_add_word(d,1)) goto err;
- }
-#endif
-
- r->neg=BN_is_zero(r)?0:m->neg;
- d->neg=m->neg^recp->N.neg;
- ret=1;
-err:
- BN_CTX_end(ctx);
- 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.
+{
+ 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.
*/
-int BN_reciprocal(BIGNUM *r, BIGNUM *m, int len, BN_CTX *ctx)
- {
- int ret= -1;
- BIGNUM t;
-
- BN_init(&t);
-
- BN_zero(&t);
- if (!BN_set_bit(&t,len)) goto err;
-
- if (!BN_div(r,NULL,&t,m,ctx)) goto err;
- ret=len;
-err:
- BN_free(&t);
- return(ret);
- }
-
+/* 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);
+}