提交 081991ac 编写于 作者: G Geoff Thorpe

With the new dynamic BN_CTX implementation, there should be no need for

additional contexts.
上级 8a85c341
......@@ -86,12 +86,10 @@ static int rsa_builtin_keygen(RSA *rsa, int bits, BIGNUM *e_value, BN_GENCB *cb)
{
BIGNUM *r0=NULL,*r1=NULL,*r2=NULL,*r3=NULL,*tmp;
int bitsp,bitsq,ok= -1,n=0;
BN_CTX *ctx=NULL,*ctx2=NULL;
BN_CTX *ctx=NULL;
ctx=BN_CTX_new();
if (ctx == NULL) goto err;
ctx2=BN_CTX_new();
if (ctx2 == NULL) goto err;
BN_CTX_start(ctx);
r0 = BN_CTX_get(ctx);
r1 = BN_CTX_get(ctx);
......@@ -183,7 +181,7 @@ static int rsa_builtin_keygen(RSA *rsa, int bits, BIGNUM *e_value, BN_GENCB *cb)
goto err;
}
*/
if (!BN_mod_inverse(rsa->d,rsa->e,r0,ctx2)) goto err; /* d */
if (!BN_mod_inverse(rsa->d,rsa->e,r0,ctx)) goto err; /* d */
/* calculate d mod (p-1) */
if (!BN_mod(rsa->dmp1,rsa->d,r1,ctx)) goto err;
......@@ -192,7 +190,7 @@ static int rsa_builtin_keygen(RSA *rsa, int bits, BIGNUM *e_value, BN_GENCB *cb)
if (!BN_mod(rsa->dmq1,rsa->d,r2,ctx)) goto err;
/* calculate inverse of q mod p */
if (!BN_mod_inverse(rsa->iqmp,rsa->q,rsa->p,ctx2)) goto err;
if (!BN_mod_inverse(rsa->iqmp,rsa->q,rsa->p,ctx)) goto err;
ok=1;
err:
......@@ -203,7 +201,6 @@ err:
}
BN_CTX_end(ctx);
BN_CTX_free(ctx);
BN_CTX_free(ctx2);
return ok;
}
......
Markdown is supported
0% .
You are about to add 0 people to the discussion. Proceed with caution.
先完成此消息的编辑!
想要评论请 注册