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authorMatt Caswell <matt@openssl.org>2015-01-22 04:40:55 +0100
committerMatt Caswell <matt@openssl.org>2015-01-22 10:20:09 +0100
commit0f113f3ee4d629ef9a4a30911b22b224772085e5 (patch)
treee014603da5aed1d0751f587a66d6e270b6bda3de /crypto/bn/bn_recp.c
parentMore tweaks for comments due indent issues (diff)
downloadopenssl-0f113f3ee4d629ef9a4a30911b22b224772085e5.tar.xz
openssl-0f113f3ee4d629ef9a4a30911b22b224772085e5.zip
Run util/openssl-format-source -v -c .
Reviewed-by: Tim Hudson <tjh@openssl.org>
Diffstat (limited to 'crypto/bn/bn_recp.c')
-rw-r--r--crypto/bn/bn_recp.c318
1 files changed, 165 insertions, 153 deletions
diff --git a/crypto/bn/bn_recp.c b/crypto/bn/bn_recp.c
index 8404086201..db5660bd13 100644
--- a/crypto/bn/bn_recp.c
+++ b/crypto/bn/bn_recp.c
@@ -5,21 +5,21 @@
* 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:
@@ -34,10 +34,10 @@
* 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
+ * 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
@@ -49,128 +49,131 @@
* 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.]
*/
-
-
#include "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;
- }
+{
+ BN_init(&(recp->N));
+ BN_init(&(recp->Nr));
+ recp->num_bits = 0;
+ recp->flags = 0;
+}
BN_RECP_CTX *BN_RECP_CTX_new(void)
- {
- BN_RECP_CTX *ret;
+{
+ BN_RECP_CTX *ret;
- if ((ret=(BN_RECP_CTX *)OPENSSL_malloc(sizeof(BN_RECP_CTX))) == NULL)
- return(NULL);
+ if ((ret = (BN_RECP_CTX *)OPENSSL_malloc(sizeof(BN_RECP_CTX))) == NULL)
+ return (NULL);
- BN_RECP_CTX_init(ret);
- ret->flags=BN_FLG_MALLOCED;
- return(ret);
- }
+ BN_RECP_CTX_init(ret);
+ 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)
- OPENSSL_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)
- {
- if (!BN_copy(&(recp->N),d)) return 0;
- BN_zero(&(recp->Nr));
- recp->num_bits=BN_num_bits(d);
- recp->shift=0;
- return(1);
- }
+{
+ 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);
- }
+ 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)) 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;
+ 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))
+ 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)))|
@@ -178,59 +181,68 @@ int BN_div_recp(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m,
* <= |(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;
+ 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;
#if 1
- 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;
- }
+ 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;
+ }
#endif
- 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->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);
- }
+{
+ 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);
+}